Skip to content
geeksforgeeks
  • Courses
    • DSA to Development
    • Get IBM Certification
    • Newly Launched!
      • Master Django Framework
      • Become AWS Certified
    • For Working Professionals
      • Interview 101: DSA & System Design
      • Data Science Training Program
      • JAVA Backend Development (Live)
      • DevOps Engineering (LIVE)
      • Data Structures & Algorithms in Python
    • For Students
      • Placement Preparation Course
      • Data Science (Live)
      • Data Structure & Algorithm-Self Paced (C++/JAVA)
      • Master Competitive Programming (Live)
      • Full Stack Development with React & Node JS (Live)
    • Full Stack Development
    • Data Science Program
    • All Courses
  • Tutorials
    • Data Structures & Algorithms
    • ML & Data Science
    • Interview Corner
    • Programming Languages
    • Web Development
    • CS Subjects
    • DevOps And Linux
    • School Learning
  • Practice
    • Build your AI Agent
    • GfG 160
    • Problem of the Day
    • Practice Coding Problems
    • GfG SDE Sheet
  • Contests
    • Accenture Hackathon (Ending Soon!)
    • GfG Weekly [Rated Contest]
    • Job-A-Thon Hiring Challenge
    • All Contests and Events
  • DSA
  • Interview Problems on DP
  • Practice DP
  • MCQs on DP
  • Tutorial on Dynamic Programming
  • Optimal Substructure
  • Overlapping Subproblem
  • Memoization
  • Tabulation
  • Tabulation vs Memoization
  • 0/1 Knapsack
  • Unbounded Knapsack
  • Subset Sum
  • LCS
  • LIS
  • Coin Change
  • Word Break
  • Egg Dropping Puzzle
  • Matrix Chain Multiplication
  • Palindrome Partitioning
  • DP on Arrays
  • DP with Bitmasking
  • Digit DP
  • DP on Trees
  • DP on Graph
Open In App
Next Article:
Palindrome Partitioning
Next article icon

Ways to arrange Balls such that adjacent balls are of different types

Last Updated : 07 Nov, 2024
Comments
Improve
Suggest changes
Like Article
Like
Report
Try it on GfG Practice
redirect icon

There are ‘p’ balls of type P, ‘q’ balls of type Q and ‘r’ balls of type R. Using the balls we want to create a straight line such that no two balls of the same type are adjacent.
Examples : 

Input: p = 1, q = 1, r = 0
Output: 2
Explanation: There are only two arrangements PQ and QP

Input: p = 1, q = 1, r = 1
Output: 6
Explanation: There are only six arrangements PQR, QPR, QRP, RQP, PRQ and RPQ

Input: p = 2, q = 1, r = 1
Output: 12
Explanation: There are twelve arrangements PQRP, PRQP, QPRP, RPQP, PQPR, PRPQ, QPQR, RQRP, PQRQ, QRQP, RPQR, and QRPR

Table of Content

  • Using Recursion – O(3^n) Time and O(n) Space
  • Using Top-Down DP (Memoization) – O(p*q*r) Time and O(p*q*r) Space
  • Using Bottom-Up DP (Tabulation) – O(p*q*r) Time and O(p*q*r) Space

Using Recursion – O(3^n) Time and O(n) Space

To solve this problem, we use a recursive approach, placing the last ball as one of three types. Each ball type is represented by a digit in the recurrence relation: 0 for P, 1 for Q, and 2 for R. If the last ball is of type P (0), the next ball must be of type Q or R, and similarly for the last balls of type Q (1) and R (2).

Recurrence Relation:

  • if the last ball is of type P: countWays(p, q, r, last) = countWays(p-1, q, r, 1) + countWays(p-1, q, r, 2)
  • if the last ball is of type Q: countWays(p, q, r, last) = countWays(p, q-1, r, 0) + countWays(p, q-1, r, 2)
  • if the last ball is of type R: countWays(p, q, r, last) = countWays(p, q, r-1, 0) + countWays(p, q-1, r-1, 1)

Base Cases:

if (p < 0), (q < 0) or (r < 0) then countWays(p, q, r, last) = 0. if only one ball ramains of specific type, then:

  • countWays(p, q, r, 0) = 1 if p=1 and q=0 and r=0
  • countWays(p, q, r, 1) = 1 if p=0 and q=1 and r=0
  • countWays(p, q, r, 2) = 1 if p=0 and q=0 and r=1
C++
// C++ program to count number of ways to arrange three // types of balls such that no two balls of same color // are adjacent to each other using recursion  #include <iostream> using namespace std;  // Returns count of arrangements where last placed ball is // 'last'.  'last' is 0 for 'p', 1 for 'q' and 2 for 'r' int countUtil(int p, int q, int r, int last) {      // if number of balls of any color becomes less     // than 0 the number of ways arrangements is 0.     if (p < 0 || q < 0 || r < 0)         return 0;      // If last ball required is of type P and the number     // of balls of P type is 1 while number of balls of     // other color is 0 the number of ways is 1.     if (p == 1 && q == 0 && r == 0 && last == 0)         return 1;      // Same case as above for 'q' and 'r'     if (p == 0 && q == 1 && r == 0 && last == 1)         return 1;     if (p == 0 && q == 0 && r == 1 && last == 2)         return 1;      // if last ball required is P and the number of ways is     // the sum of number of ways to form sequence with 'p-1' P     // balls, q Q Balls and r R balls ending with Q and R.     if (last == 0)         return countUtil(p - 1, q, r, 1) + countUtil(p - 1, q, r, 2);      // Same as above case for 'q' and 'r'     if (last == 1)         return countUtil(p, q - 1, r, 0) + countUtil(p, q - 1, r, 2);     if (last == 2)         return countUtil(p, q, r - 1, 0) + countUtil(p, q, r - 1, 1); }  // Returns count of required arrangements int countWays(int p, int q, int r) {          // Three cases arise:     // Last required balls is type P     // Last required balls is type Q     // Last required balls is type R     return countUtil(p, q, r, 0) + countUtil(p, q, r, 1) + countUtil(p, q, r, 2); }  int main() {     int p = 1, q = 1, r = 1;     int res = countWays(p, q, r);     cout << res << endl;      return 0; } 
Java
// Java program to count number // of ways to arrange three types of // balls such that no two balls of // same color are adjacent to each other // using recursion  import java.io.*; import java.util.*;  class GfG {      // Returns count of arrangements     // where last placed ball is     // 'last'. 'last' is 0 for 'p',     // 1 for 'q' and 2 for 'r'     static int countUtil(int p, int q, int r, int last) {                // if number of balls of any         // color becomes less than 0         // the number of ways arrangements is 0.         if (p < 0 || q < 0 || r < 0)             return 0;          // If last ball required is         // of type P and the number         // of balls of P type is 1         // while number of balls of         // other color is 0 the number         // of ways is 1.         if (p == 1 && q == 0 && r == 0 && last == 0)             return 1;          // Same case as above for 'q' and 'r'         if (p == 0 && q == 1 && r == 0 && last == 1)             return 1;         if (p == 0 && q == 0 && r == 1 && last == 2)             return 1;          // if last ball required is P         // and the number of ways is         // the sum of number of ways         // to form sequence with 'p-1' P         // balls, q Q Balls and r R balls         // ending with Q and R.         if (last == 0)             return countUtil(p - 1, q, r, 1)                 + countUtil(p - 1, q, r, 2);          // Same as above case for 'q' and 'r'         if (last == 1)             return countUtil(p, q - 1, r, 0)                 + countUtil(p, q - 1, r, 2);          if (last == 2)             return countUtil(p, q, r - 1, 0)                 + countUtil(p, q, r - 1, 1);          return 0;     }      // Returns count of required arrangements     static int countWays(int p, int q, int r) {          // Three cases arise:         // Last required balls is type P         // Last required balls is type Q         // Last required balls is type R         return countUtil(p, q, r, 0) + countUtil(p, q, r, 1)             + countUtil(p, q, r, 2);     }      public static void main(String[] args) {         int p = 1, q = 1, r = 1;         int res = countWays(p, q, r);          System.out.print(res);     } } 
Python
# Python3 program to count # number of ways to arrange # three types of balls such # that no two balls of same # color are adjacent to each # other using recursion  # Returns count of arrangements # where last placed ball is # 'last'. 'last' is 0 for 'p', # 1 for 'q' and 2 for 'r'   def countUtil(p, q, r, last):      # if number of balls of     # any color becomes less     # than 0 the number of     # ways arrangements is 0.     if (p < 0 or q < 0 or r < 0):         return 0      # If last ball required is     # of type P and the number     # of balls of P type is 1     # while number of balls of     # other color is 0 the number     # of ways is 1.     if (p == 1 and q == 0             and r == 0 and last == 0):         return 1      # Same case as above     # for 'q' and 'r'     if (p == 0 and q == 1             and r == 0 and last == 1):         return 1      if (p == 0 and q == 0 and             r == 1 and last == 2):         return 1      # if last ball required is P     # and the number of ways is     # the sum of number of ways     # to form sequence with 'p-1' P     # balls, q Q Balls and r R     # balls ending with Q and R.     if (last == 0):         return (countUtil(p - 1, q, r, 1) +                 countUtil(p - 1, q, r, 2));      # Same as above case     # for 'q' and 'r'     if (last == 1):         return (countUtil(p, q - 1, r, 0) +                 countUtil(p, q - 1, r, 2));     if (last == 2):         return (countUtil(p, q, r - 1, 0) +                 countUtil(p, q, r - 1, 1));   def countWays(p, q, r):      # Three cases arise:     # Last required balls is type P     # Last required balls is type Q     # Last required balls is type R     return (countUtil(p, q, r, 0) +             countUtil(p, q, r, 1) +             countUtil(p, q, r, 2));   p = 1 q = 1 r = 1 res = countWays(p, q, r) print(res) 
C#
// C# program to count number // of ways to arrange three types of // balls such that no two balls of // same color are adjacent to each other // uisng recursion  using System;  class GfG {      // Returns count of arrangements     // where last placed ball is     // 'last'. 'last' is 0 for 'p',     // 1 for 'q' and 2 for 'r'     static int countUtil(int p, int q, int r, int last) {          // if number of balls of any         // color becomes less than 0         // the number of ways         // arrangements is 0.         if (p < 0 || q < 0 || r < 0)             return 0;          // If last ball required is         // of type P and the number         // of balls of P type is 1         // while number of balls of         // other color is 0 the number         // of ways is 1.         if (p == 1 && q == 0 && r == 0 && last == 0)             return 1;          // Same case as above for 'q' and 'r'         if (p == 0 && q == 1 && r == 0 && last == 1)             return 1;         if (p == 0 && q == 0 && r == 1 && last == 2)             return 1;          // if last ball required is P         // and the number of ways is         // the sum of number of ways         // to form sequence with 'p-1' P         // balls, q Q Balls and r R balls         // ending with Q and R.         if (last == 0)             return countUtil(p - 1, q, r, 1)                 + countUtil(p - 1, q, r, 2);          // Same as above case for 'q' and 'r'         if (last == 1)             return countUtil(p, q - 1, r, 0)                 + countUtil(p, q - 1, r, 2);          if (last == 2)             return countUtil(p, q, r - 1, 0)                 + countUtil(p, q, r - 1, 1);          return 0;     }      // Returns count of required arrangements     static int countWays(int p, int q, int r) {          // Three cases arise:         // 1. Last required balls is type P         // 2. Last required balls is type Q         // 3. Last required balls is type R         return countUtil(p, q, r, 0) + countUtil(p, q, r, 1)             + countUtil(p, q, r, 2);     }      static void Main() {         int p = 1, q = 1, r = 1;         int res = countWays(p, q, r);          Console.Write(res);     } } 
JavaScript
// JavaScript program to count number // of ways to arrange three // types of balls such that no // two balls of same color // are adjacent to each other // using recursion  // Returns count of arrangements // where last placed ball is // 'last'. 'last' is 0 for 'p', // 1 for 'q' and 2 for 'r' function countUtil(p, q, r, last) {      // if number of balls of any     // color becomes less than 0     // the number of ways arrangements is 0.     if (p < 0 || q < 0 || r < 0)         return 0;      // If last ball required is     // of type P and the number     // of balls of P type is 1     // while number of balls of     // other color is 0 the number     // of ways is 1.     if (p == 1 && q == 0 && r == 0 && last == 0)         return 1;      // Same case as above for 'q' and 'r'     if (p == 0 && q == 1 && r == 0 && last == 1)         return 1;     if (p == 0 && q == 0 && r == 1 && last == 2)         return 1;      // if last ball required is P     // and the number of ways is     // the sum of number of ways     // to form sequence with 'p-1' P     // balls, q Q Balls and r R balls     // ending with Q and R.     if (last == 0)         return countUtil(p - 1, q, r, 1)                + countUtil(p - 1, q, r, 2);      // Same as above case for 'q' and 'r'     if (last == 1)         return countUtil(p, q - 1, r, 0)                + countUtil(p, q - 1, r, 2);      if (last == 2)         return countUtil(p, q, r - 1, 0)                + countUtil(p, q, r - 1, 1);      return 0; }  // Returns count of required arrangements function countWays(p, q, r) {          return countUtil(p, q, r, 0) + countUtil(p, q, r, 1)     							 + countUtil(p, q, r, 2); }  let p = 1, q = 1, r = 1; let res = countWays(p, q, r); console.log(res); 

Output
6 

Using Top-Down DP (Memoization) – O(p*q*r) Time and O(p*q*r) Space

If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming.

1. Optimal Substructure:

The solution to the problem of arranging balls with no two adjacent balls of the same type can be derived from the optimal solutions of smaller subproblems.

2. Overlapping Subproblems:

In the recursive solution, we notice that many subproblems are solved multiple times. For example, when calculating countWays(p, q, r, last), the same values of p, q, r are computed multiple times.

  • The recursive solution involves changing four parameters: the number of remaining balls of each type (p, q, r) and the last placed ball type (last). We need to track all these parameters, so we create a 4D array of size (p+1) x (q+1) x (r+1) x 3. This is because the values of p, q, and r will be in the range [0, p], [0, q], and [0, r], respectively, and the value of last will range from 0 to 2 (representing the three ball types: P, Q, and R).
  • We initialize this 4D array with -1 to indicate that no subproblems have been computed yet.
  • Before computing the number of arrangements for any subproblem, we check if the value at memo[p][q][r][last] is -1. If it is, we proceed to compute the result using the recursive relations. If it is, we proceed to compute the result. otherwise, we return the stored result.
C++
// C++ program to count number of ways to arrange three // types of balls such that no two balls of same color // are adjacent to each other using tabulation #include <iostream> #include <vector> using namespace std;  // Returns count of arrangements where last placed ball is // 'last'. 'last' is 0 for 'p', 1 for 'q' and 2 for 'r' int countUtil(int p, int q, int r, int last,                vector<vector<vector<vector<int>>>> &memo) {        // If number of balls of any color becomes less than 0,     // the number of ways to arrange them is 0.     if (p < 0 || q < 0 || r < 0)         return 0;      // Base cases: when only one ball is left, return 1 if    	// the last ball is the right type     if (p == 1 && q == 0 && r == 0 && last == 0)         return 1;     if (p == 0 && q == 1 && r == 0 && last == 1)         return 1;     if (p == 0 && q == 0 && r == 1 && last == 2)         return 1;      // If this subproblem is already evaluated (memoized),   	// return the stored result     if (memo[p][q][r][last] != -1)         return memo[p][q][r][last];      // Recursive calls to calculate number of arrangements   	// based on the last ball     if (last == 0)                // If the last ball is P, the next ball can be Q or R         memo[p][q][r][last] = countUtil(p - 1, q, r, 1, memo)       						+ countUtil(p - 1, q, r, 2, memo);     else if (last == 1)                // If the last ball is Q, the next ball can be P or R         memo[p][q][r][last] = countUtil(p, q - 1, r, 0, memo)       						+ countUtil(p, q - 1, r, 2, memo);     else                // If the last ball is R, the next ball can be P or Q         memo[p][q][r][last] = countUtil(p, q, r - 1, 0, memo)        						+ countUtil(p, q, r - 1, 1, memo);      return memo[p][q][r][last]; }  // Wrapper function to initialize memoization table // and call countUtil int countWays(int p, int q, int r) {      // Create a 4D vector for memoization with size   	// (p+1) x (q+1) x (r+1) x 3     vector<vector<vector<vector<int>>>> memo(         p + 1,         vector<vector<vector<int>>>(             q + 1, vector<vector<int>>(                        r + 1, vector<int>(3, -1))));      // Call countUtil for all possible last ball types (0, 1, 2)     int ans = countUtil(p, q, r, 0, memo) +      countUtil(p, q, r, 1, memo) + countUtil(p, q, r, 2, memo);     return ans; }  int main() {      int p = 1, q = 1, r = 1;     int res = countWays(p, q, r);     cout << res << endl;      return 0; } 
Java
// Java program to count number // of ways to arrange three // types of balls such that no // two balls of same color // are adjacent to each other import java.util.*;  class GfG {      // Returns count of arrangements where last placed ball     // is 'last'. 'last' is 0 for 'p', 1 for 'q' and 2 for     // 'r'     static int     countUtil(int p, int q, int r, int last,               List<List<List<List<Integer> > > > memo) {                // If number of balls of any color becomes less than         // 0, the number of ways to arrange them is 0.         if (p < 0 || q < 0 || r < 0)             return 0;          // Base cases: when only one ball is left, return 1         // if the last ball is the right type         if (p == 1 && q == 0 && r == 0 && last == 0)             return 1;         if (p == 0 && q == 1 && r == 0 && last == 1)             return 1;         if (p == 0 && q == 0 && r == 1 && last == 2)             return 1;          // If this subproblem is already evaluated         // (memoized), return the stored result         if (memo.get(p).get(q).get(r).get(last) != -1)             return memo.get(p).get(q).get(r).get(last);          // Recursive calls to calculate number of         // arrangements based on the last ball         if (last == 0)                        // If the last ball is P, the next ball can be Q             // or R             memo.get(p).get(q).get(r).set(                 last,                 countUtil(p - 1, q, r, 1, memo)                     + countUtil(p - 1, q, r, 2, memo));         else if (last == 1)                        // If the last ball is Q, the next ball can be P             // or R             memo.get(p).get(q).get(r).set(                 last,                 countUtil(p, q - 1, r, 0, memo)                     + countUtil(p, q - 1, r, 2, memo));         else                        // If the last ball is R, the next ball can be P             // or Q             memo.get(p).get(q).get(r).set(                 last,                 countUtil(p, q, r - 1, 0, memo)                     + countUtil(p, q, r - 1, 1, memo));          return memo.get(p).get(q).get(r).get(last);     }      // Wrapper function to initialize memoization table and     // call countUtil     static int countWays(int p, int q, int r) {          // Create a 4D list for memoization with size (p+1)         // x (q+1) x (r+1) x 3         List<List<List<List<Integer> > > > memo             = new ArrayList<>();          for (int i = 0; i <= p; i++) {             List<List<List<Integer> > > level2                 = new ArrayList<>();             for (int j = 0; j <= q; j++) {                 List<List<Integer> > level3                     = new ArrayList<>();                 for (int k = 0; k <= r; k++) {                     List<Integer> level4 = new ArrayList<>(                         Arrays.asList(-1, -1, -1));                     level3.add(level4);                 }                 level2.add(level3);             }             memo.add(level2);         }          // Call countUtil for all possible last ball types         // (0, 1, 2)         int ans = countUtil(p, q, r, 0, memo)                   + countUtil(p, q, r, 1, memo)                   + countUtil(p, q, r, 2, memo);         return ans;     }      public static void main(String[] args) {         int p = 1, q = 1, r = 1;         int res = countWays(p, q, r);          System.out.print(res);     } } 
Python
# Python program to count number of ways to arrange three # types of balls such that no two balls of same color # are adjacent to each other using tabulation   def countUtil(p, q, r, last, memo):      # If number of balls of any color becomes less than 0,     # the number of ways to arrange them is 0.     if p < 0 or q < 0 or r < 0:         return 0      # Base cases: when only one ball is left, return 1 if     # the last ball is the right type     if p == 1 and q == 0 and r == 0 and last == 0:         return 1     if p == 0 and q == 1 and r == 0 and last == 1:         return 1     if p == 0 and q == 0 and r == 1 and last == 2:         return 1      # If this subproblem is already evaluated      # (memoized), return the stored result     if memo[p][q][r][last] != -1:         return memo[p][q][r][last]      # Recursive calls to calculate number of      # arrangements based on the last ball     if last == 0:                # If the last ball is P, the next ball can be Q or R         memo[p][q][r][last] = countUtil(             p - 1, q, r, 1, memo) + countUtil(p - 1, q, r, 2, memo)     elif last == 1:                # If the last ball is Q, the next ball can be P or R         memo[p][q][r][last] = countUtil(             p, q - 1, r, 0, memo) + countUtil(p, q - 1, r, 2, memo)     else:                # If the last ball is R, the next ball can be P or Q         memo[p][q][r][last] = countUtil(             p, q, r - 1, 0, memo) + countUtil(p, q, r - 1, 1, memo)      return memo[p][q][r][last]   # Wrapper function to initialize memoization  # table and call countUtil def countWays(p, q, r):      # Create a 4D list for memoization with size     # (p+1) x (q+1) x (r+1) x 3     memo = [[[[-1 for _ in range(3)] for _ in range(r + 1)]              for _ in range(q + 1)] for _ in range(p + 1)]      # Call countUtil for all possible last     # ball types (0, 1, 2)     ans = countUtil(p, q, r, 0, memo) + countUtil(p, q, r,                                                   1, memo) + countUtil(p, q, r, 2, memo)     return ans   if __name__ == "__main__":      p = 1     q = 1     r = 1      res = countWays(p, q, r)     print(res) 
C#
// C# program to count number // of ways to arrange three // types of balls such that no // two balls of same color // are adjacent to each other using System;  class GfG {      // Returns count of arrangements where last placed ball     // is 'last'. 'last' is 0 for 'p', 1 for 'q', and 2 for     // 'r'     static int countUtil(int p, int q, int r, int last,                          int[, , , ] memo) {          // If number of balls of any color becomes less than         // 0, the number of ways to arrange them is 0.         if (p < 0 || q < 0 || r < 0)             return 0;          // Base cases: when only one ball of each type is         // left         if (p == 1 && q == 0 && r == 0 && last == 0)             return 1;         if (p == 0 && q == 1 && r == 0 && last == 1)             return 1;         if (p == 0 && q == 0 && r == 1 && last == 2)             return 1;          // If this subproblem is already evaluated         // (memoized), return the stored result         if (memo[p, q, r, last] != -1)             return memo[p, q, r, last];          // Recursive calls to calculate number of         // arrangements based on the last ball         if (last == 0) {              // If the last ball is P, the next ball can be Q             // or R             memo[p, q, r, last]                 = countUtil(p - 1, q, r, 1, memo)                   + countUtil(p - 1, q, r, 2, memo);         }         else if (last == 1) {              // If the last ball is Q, the next ball can be P             // or R             memo[p, q, r, last]                 = countUtil(p, q - 1, r, 0, memo)                   + countUtil(p, q - 1, r, 2, memo);         }         else {              // If the last ball is R, the next ball can be P             // or Q             memo[p, q, r, last]                 = countUtil(p, q, r - 1, 0, memo)                   + countUtil(p, q, r - 1, 1, memo);         }          // Return the computed result and store it in the         // memoization table         return memo[p, q, r, last];     }      // Wrapper function to initialize memoization table and     // call countUtil     static int countWays(int p, int q, int r) {                // Create a 4D array for memoization with size (p+1)         // x (q+1) x (r+1) x 3         int[, , , ] memo = new int[p + 1, q + 1, r + 1, 3];          // Initialize all elements of the memo array to -1         for (int i = 0; i <= p; i++) {             for (int j = 0; j <= q; j++) {                 for (int k = 0; k <= r; k++) {                     for (int l = 0; l < 3; l++) {                         memo[i, j, k, l] = -1;                     }                 }             }         }          // Call countUtil for all possible last ball types         // (0, 1, 2)         int ans = countUtil(p, q, r, 0, memo)                   + countUtil(p, q, r, 1, memo)                   + countUtil(p, q, r, 2, memo);         return ans;     }      static void Main() {         int p = 1, q = 1, r = 1;         int res = countWays(p, q, r);         Console.WriteLine(res);     } } 
JavaScript
// JavaScript program to count number of ways to arrange // three types of balls such that no two balls of same color // are adjacent to each other using tabulation  // Returns count of arrangements where last placed ball is // 'last'. 'last' is 0 for 'p', 1 for 'q' and 2 for 'r' function countUtil(p, q, r, last, memo) {      // If number of balls of any color becomes less than 0,     // the number of ways to arrange them is 0.     if (p < 0 || q < 0 || r < 0) {         return 0;     }      // Base cases: when only one ball is left, return 1 if     // the last ball is the right type     if (p === 1 && q === 0 && r === 0 && last === 0) {         return 1;     }     if (p === 0 && q === 1 && r === 0 && last === 1) {         return 1;     }     if (p === 0 && q === 0 && r === 1 && last === 2) {         return 1;     }      // If this subproblem is already evaluated (memoized),     // return the stored result     if (memo[p][q][r][last] !== -1) {         return memo[p][q][r][last];     }      // Recursive calls to calculate number of arrangements     // based on the last ball     if (last === 0) {              // If the last ball is P, the next ball can be Q or         // R         memo[p][q][r][last]             = countUtil(p - 1, q, r, 1, memo)               + countUtil(p - 1, q, r, 2, memo);     }     else if (last === 1) {              // If the last ball is Q, the next ball can be P or         // R         memo[p][q][r][last]             = countUtil(p, q - 1, r, 0, memo)               + countUtil(p, q - 1, r, 2, memo);     }     else {              // If the last ball is R, the next ball can be P or         // Q         memo[p][q][r][last]             = countUtil(p, q, r - 1, 0, memo)               + countUtil(p, q, r - 1, 1, memo);     }      return memo[p][q][r][last]; }  // Wrapper function to initialize memoization  // table and call countUtil function countWays(p, q, r) {      // Create a 4D array for memoization with size (p+1) x     // (q+1) x (r+1) x 3     let memo = new Array(p + 1);     for (let i = 0; i <= p; i++) {         memo[i] = new Array(q + 1);         for (let j = 0; j <= q; j++) {             memo[i][j] = new Array(r + 1);             for (let k = 0; k <= r; k++) {                 memo[i][j][k] = new Array(3).fill(                     -1);             }         }     }      // Call countUtil for all possible last ball types (0,     // 1, 2)     let ans = countUtil(p, q, r, 0, memo)               + countUtil(p, q, r, 1, memo)               + countUtil(p, q, r, 2, memo);     return ans; }  let p = 1, q = 1, r = 1; let res = countWays(p, q, r); console.log(res); 

Output
6 

Using Bottom-Up DP (Tabulation) – O(p*q*r) Time and O(p*q*r) Space

The approach is similar to the previous one. just instead of breaking down the problem recursively, we iteratively build up the solution by calculating in bottom-up manner.

So we will create 4D array of size (p+1)*(q+1)*(r+1)*last. and the state dp[i][j][k][l] will represent the number of ways to arrange the balls such that:

  • i is the count of balls of type P placed so far.
  • j is the count of balls of type Q placed so far.
  • k is the count of balls of type R placed so far.
  • l is the type of the last ball placed (l = 0 for P, l = 1 for Q, and l = 2 for R).

1. For the case where the last ball placed is of type P (i.e., l=0): if i>0, then dp[i][j][k][0] = dp[i-1][j][k][1] + dp[i-1][j][k][2]

2. For the case where the last ball placed is of type Q (i.e. ,l=1): if j>0, then dp[i][j][k][1]=dp[i][j-1][k][0]+dp[i][j-1][k][2]

3. For the case where the last ball placed is of type R (i.e., l=2): if k>0, then dp[i][j][k][2]=dp[i][j][k-1][0]+dp[i][j][k-1][1]

Base Cases

  • dp[1][0][0][0]=1 (Only one P ball left, and the last placed ball is P)
  • dp[0][1][0][1]=1 (Only one Q ball left, and the last placed ball is Q)
  • dp[0][0][1][2]=1 (Only one R ball left, and the last placed ball is R)
C++
// c++ program to count number of ways to arrange three // types of balls such that no two balls of same color // are adjacent to each other using tabulation  #include <bits/stdc++.h> using namespace std;  // Function to count the number of arrangements int countWays(int p, int q, int r) {      // Create a 4D DP table (p+1) x (q+1) x (r+1) x 3     vector<vector<vector<vector<int>>>> dp(         p + 1, vector<vector<vector<int>>>(q + 1,  		vector<vector<int>>(r + 1, vector<int>(3, 0))));      // Base cases for when only one ball of each type is left     // Only p left and the last ball is p     dp[1][0][0][0] = 1;      // Only q left and the last ball is q     dp[0][1][0][1] = 1;      // Only r left and the last ball is r     dp[0][0][1][2] = 1;      // Iteratively fill the DP table     for (int pCount = 0; pCount <= p; pCount++) {         for (int qCount = 0; qCount <= q; qCount++) {             for (int rCount = 0; rCount <= r; rCount++) {                 for (int last = 0; last < 3; last++) {                      // If the count of balls is zero, skip                     if (pCount == 0 && qCount == 0 && rCount == 0)                         continue;                      if (last == 0) {                          // Last ball was P, so next can be Q or R                         if (pCount > 0)                             dp[pCount][qCount][rCount][last]                           	+= dp[pCount - 1][qCount][rCount][1];                         if (pCount > 0)                             dp[pCount][qCount][rCount][last]                            	+= dp[pCount - 1][qCount][rCount][2];                     }                     else if (last == 1) {                          // Last ball was Q, so next can be P or R                         if (qCount > 0) {                             dp[pCount][qCount][rCount][last]                              += dp[pCount][qCount - 1][rCount][0];                             dp[pCount][qCount][rCount][last]                              += dp[pCount][qCount - 1][rCount][2];                         }                     }                     else {                          // Last ball was R, so next can be P or Q                         if (rCount > 0) {                             dp[pCount][qCount][rCount][last]                              += dp[pCount][qCount][rCount - 1][0];                             dp[pCount][qCount][rCount][last]                              += dp[pCount][qCount][rCount - 1][1];                         }                     }                 }             }         }     }      // The answer is the sum of all configurations   	// for the given p, q, r     int ans = dp[p][q][r][0]      + dp[p][q][r][1] + dp[p][q][r][2];     return ans; }  int main() {     int p = 1, q = 1, r = 1;     int res = countWays(p, q, r);     cout << res << endl;     return 0; } 
Java
// java program to count number of ways to arrange three // types of balls such that no two balls of same color // are adjacent to each other using tabulation  import java.util.*;  class GfG {      // Function to count the number of arrangements     // Function to count the number of arrangements     static int countWays(int p, int q, int r) {          // Create a 4D DP table (p+1) x (q+1) x (r+1) x 3         // using arrays         int[][][][] dp = new int[p + 1][q + 1][r + 1][3];          // Base cases for when only one ball of each type is         // left Only p left and the last ball is p         dp[1][0][0][0] = 1;          // Only q left and the last ball is q         dp[0][1][0][1] = 1;          // Only r left and the last ball is r         dp[0][0][1][2] = 1;          // Iteratively fill the DP table         for (int pCount = 0; pCount <= p; pCount++) {             for (int qCount = 0; qCount <= q; qCount++) {                 for (int rCount = 0; rCount <= r;                      rCount++) {                     for (int last = 0; last < 3; last++) {                                                // If the count of balls is zero,                         // skip                         if (pCount == 0 && qCount == 0                             && rCount == 0)                             continue;                          if (last == 0) {                                                        // Last ball was P, so next can                             // be Q or R                             if (pCount > 0)                                 dp[pCount][qCount][rCount]                                   [last]                                     += dp[pCount - 1]                                          [qCount][rCount]                                          [1];                             if (pCount > 0)                                 dp[pCount][qCount][rCount]                                   [last]                                     += dp[pCount - 1]                                          [qCount][rCount]                                          [2];                         }                         else if (last == 1) {                                                        // Last ball was Q, so next can                             // be P or R                             if (qCount > 0) {                                 dp[pCount][qCount][rCount]                                   [last]                                     += dp[pCount]                                          [qCount - 1]                                          [rCount][0];                                 dp[pCount][qCount][rCount]                                   [last]                                     += dp[pCount]                                          [qCount - 1]                                          [rCount][2];                             }                         }                         else {                                                        // Last ball was R, so next can                             // be P or Q                             if (rCount > 0) {                                 dp[pCount][qCount][rCount]                                   [last]                                     += dp[pCount][qCount]                                          [rCount - 1][0];                                 dp[pCount][qCount][rCount]                                   [last]                                     += dp[pCount][qCount]                                          [rCount - 1][1];                             }                         }                     }                 }             }         }          // The answer is the sum of all configurations for         // the given p, q, r         int ans = dp[p][q][r][0] + dp[p][q][r][1]                   + dp[p][q][r][2];         return ans;     }      public static void main(String[] args) {         int p = 1, q = 1, r = 1;         System.out.println(countWays(p, q, r));     } } 
Python
# python program to count number of ways to arrange three # types of balls such that no two balls of same color # are adjacent to each other using tabulation   def countWays(p, q, r):        # Create a 4D DP table (p+1) x (q+1) x (r+1) x 3 using lists     dp = [[[[0 for _ in range(3)] for _ in range(r + 1)]            for _ in range(q + 1)] for _ in range(p + 1)]      # Base cases for when only one ball of each type is left     # Only p left and the last ball is p     dp[1][0][0][0] = 1      # Only q left and the last ball is q     dp[0][1][0][1] = 1      # Only r left and the last ball is r     dp[0][0][1][2] = 1      # Iteratively fill the DP table     for pCount in range(p + 1):         for qCount in range(q + 1):             for rCount in range(r + 1):                 for last in range(3):                                        # If the count of balls is zero, skip                     if pCount == 0 and qCount == 0 and rCount == 0:                         continue                      if last == 0:                                                # Last ball was P, so next can be Q or R                         if pCount > 0:                             dp[pCount][qCount][rCount][last] += dp[pCount -                                                                    1][qCount][rCount][1]                         if pCount > 0:                             dp[pCount][qCount][rCount][last] += dp[pCount -                                                                    1][qCount][rCount][2]                     elif last == 1:                                                # Last ball was Q, so next can be P or R                         if qCount > 0:                             dp[pCount][qCount][rCount][last] += dp[pCount][qCount - 1][rCount][0]                             dp[pCount][qCount][rCount][last] += dp[pCount][qCount - 1][rCount][2]                     else:                          # Last ball was R, so next can be P or Q                         if rCount > 0:                             dp[pCount][qCount][rCount][last] += dp[pCount][qCount][rCount - 1][0]                             dp[pCount][qCount][rCount][last] += dp[pCount][qCount][rCount - 1][1]      # The answer is the sum of all configurations     # for the given p, q, r     ans = dp[p][q][r][0] + dp[p][q][r][1] + dp[p][q][r][2]     return ans   if __name__ == "__main__":     p, q, r = 1, 1, 1     print(countWays(p, q, r)) 
C#
// c# program to count number of ways to arrange three // types of balls such that no two balls of same color // are adjacent to each other using tabulation  using System;  class GfG {      // Function to count the number of arrangements     static int countWays(int p, int q, int r) {                // Create a 4D DP table (p+1) x (q+1) x (r+1) x 3         // using arrays         int[, , , ] dp = new int[p + 1, q + 1, r + 1, 3];          // Base cases for when only one ball of each type is         // left Only p left and the last ball is p         dp[1, 0, 0, 0] = 1;          // Only q left and the last ball is q         dp[0, 1, 0, 1] = 1;          // Only r left and the last ball is r         dp[0, 0, 1, 2] = 1;          // Iteratively fill the DP table         for (int pCount = 0; pCount <= p; pCount++) {             for (int qCount = 0; qCount <= q; qCount++) {                 for (int rCount = 0; rCount <= r;                      rCount++) {                     for (int last = 0; last < 3; last++) {                                                // If the count of balls is zero,                         // skip                         if (pCount == 0 && qCount == 0                             && rCount == 0)                             continue;                          if (last == 0) {                                                        // Last ball was P, so next can                             // be Q or R                             if (pCount > 0)                                 dp[pCount, qCount, rCount,                                    last]                                     += dp[pCount - 1,                                           qCount, rCount,                                           1];                             if (pCount > 0)                                 dp[pCount, qCount, rCount,                                    last]                                     += dp[pCount - 1,                                           qCount, rCount,                                           2];                         }                         else if (last == 1) {                                                        // Last ball was Q, so next can                             // be P or R                             if (qCount > 0) {                                 dp[pCount, qCount, rCount,                                    last]                                     += dp[pCount,                                           qCount - 1,                                           rCount, 0];                                 dp[pCount, qCount, rCount,                                    last]                                     += dp[pCount,                                           qCount - 1,                                           rCount, 2];                             }                         }                         else {                                                        // Last ball was R, so next can                             // be P or Q                             if (rCount > 0) {                                 dp[pCount, qCount, rCount,                                    last]                                     += dp[pCount, qCount,                                           rCount - 1, 0];                                 dp[pCount, qCount, rCount,                                    last]                                     += dp[pCount, qCount,                                           rCount - 1, 1];                             }                         }                     }                 }             }         }          // The answer is the sum of all configurations for         // the given p, q, r         int ans = dp[p, q, r, 0] + dp[p, q, r, 1]                   + dp[p, q, r, 2];         return ans;     }      static void Main() {         int p = 1, q = 1, r = 1;         int ans = countWays(p, q, r);         Console.WriteLine(ans);     } } 
JavaScript
// JavaScript program to count number of ways to arrange // three types of balls such that no two balls of same color // are adjacent to each other using tabulation  function countWays(p, q, r) {      let dp = new Array(p + 1);     for (let i = 0; i <= p; i++) {         dp[i] = new Array(q + 1);         for (let j = 0; j <= q; j++) {             dp[i][j] = new Array(r + 1);             for (let k = 0; k <= r; k++) {                 dp[i][j][k] = new Array(3).fill(0);             }         }     }      // Base cases for when only one ball of each type is     // left     dp[1][0][0][0] = 1;     dp[0][1][0][1] = 1;     dp[0][0][1][2] = 1;      // Iteratively fill the DP table     for (let pCount = 0; pCount <= p; pCount++) {         for (let qCount = 0; qCount <= q; qCount++) {             for (let rCount = 0; rCount <= r; rCount++) {                 for (let last = 0; last < 3; last++) {                      // If the count of balls is zero, skip                     if (pCount === 0 && qCount === 0                         && rCount === 0)                         continue;                      if (last === 0) {                          // Last ball was P, so next can be Q                         // or R                         if (pCount > 0)                             dp[pCount][qCount][rCount][last]                                 += dp[pCount                                       - 1][qCount][rCount][1];                         if (pCount > 0)                             dp[pCount][qCount][rCount][last]                                 += dp[pCount                                       - 1][qCount][rCount][2];                     }                     else if (last === 1) {                          // Last ball was Q, so next can be P                         // or R                         if (qCount > 0) {                             dp[pCount][qCount][rCount][last]                                 += dp[pCount][qCount                                               - 1][rCount][0];                             dp[pCount][qCount][rCount][last]                                 += dp[pCount][qCount                                               - 1][rCount][2];                         }                     }                     else {                          // Last ball was R, so next can be P                         // or Q                         if (rCount > 0) {                             dp[pCount][qCount][rCount][last]                                 += dp[pCount][qCount][rCount                                                       - 1][0];                             dp[pCount][qCount][rCount][last]                                 += dp[pCount][qCount][rCount                                                       - 1][1];                         }                     }                 }             }         }     }      // The answer is the sum of all configurations for the     // given p, q, r     let ans         = dp[p][q][r][0] + dp[p][q][r][1] + dp[p][q][r][2];     return ans; }  const p = 1, q = 1, r = 1; let ans = countWays(p, q, r); console.log(ans); 

Output
6 


Next Article
Palindrome Partitioning

B

Bhavuk Chawla
Improve
Article Tags :
  • DSA
  • Dynamic Programming
  • Mathematical
Practice Tags :
  • Dynamic Programming
  • Mathematical

Similar Reads

  • Dynamic Programming or DP
    Dynamic Programming is an algorithmic technique with the following properties. It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
    3 min read
  • What is Memoization? A Complete Tutorial
    In this tutorial, we will dive into memoization, a powerful optimization technique that can drastically improve the performance of certain algorithms. Memoization helps by storing the results of expensive function calls and reusing them when the same inputs occur again. This avoids redundant calcula
    6 min read
  • Dynamic Programming (DP) Introduction
    Dynamic Programming is a commonly used algorithmic technique used to optimize recursive solutions when same subproblems are called again. The core idea behind DP is to store solutions to subproblems so that each is solved only once. To solve DP problems, we first write a recursive solution in a way
    15+ min read
  • Tabulation vs Memoization
    Tabulation and memoization are two techniques used to implement dynamic programming. Both techniques are used when there are overlapping subproblems (the same subproblem is executed multiple times). Below is an overview of two approaches. Memoization:Top-down approachStores the results of function c
    9 min read
  • Optimal Substructure Property in Dynamic Programming | DP-2
    The following are the two main properties of a problem that suggest that the given problem can be solved using Dynamic programming: 1) Overlapping Subproblems 2) Optimal Substructure We have already discussed the Overlapping Subproblem property. Let us discuss the Optimal Substructure property here.
    4 min read
  • Overlapping Subproblems Property in Dynamic Programming | DP-1
    Dynamic Programming is an algorithmic paradigm that solves a given complex problem by breaking it into subproblems using recursion and storing the results of subproblems to avoid computing the same results again. Following are the two main properties of a problem that suggests that the given problem
    10 min read
  • Steps to solve a Dynamic Programming Problem
    Steps to solve a Dynamic programming problem:Identify if it is a Dynamic programming problem.Decide a state expression with the Least parameters.Formulate state and transition relationship.Apply tabulation or memorization.Step 1: How to classify a problem as a Dynamic Programming Problem? Typically,
    14 min read
  • Advanced Topics

    • Count Ways To Assign Unique Cap To Every Person
      Given n people and 100 types of caps labelled from 1 to 100, along with a 2D integer array caps where caps[i] represents the list of caps preferred by the i-th person, the task is to determine the number of ways the n people can wear different caps. Example: Input: caps = [[3, 4], [4, 5], [5]] Outpu
      15+ min read

    • Digit DP | Introduction
      Prerequisite : How to solve a Dynamic Programming Problem ?There are many types of problems that ask to count the number of integers 'x' between two integers say 'a' and 'b' such that x satisfies a specific property that can be related to its digits.So, if we say G(x) tells the number of such intege
      14 min read

    • Sum over Subsets | Dynamic Programming
      Prerequisite: Basic Dynamic Programming, Bitmasks Consider the following problem where we will use Sum over subset Dynamic Programming to solve it. Given an array of 2n integers, we need to calculate function F(x) = ?Ai such that x&i==i for all x. i.e, i is a bitwise subset of x. i will be a bit
      10 min read

    Easy problems in Dynamic programming

    • Coin Change - Count Ways to Make Sum
      Given an integer array of coins[] of size n representing different types of denominations and an integer sum, the task is to count all combinations of coins to make a given value sum. Note: Assume that you have an infinite supply of each type of coin. Examples: Input: sum = 4, coins[] = [1, 2, 3]Out
      15+ min read

    • Subset Sum Problem
      Given an array arr[] of non-negative integers and a value sum, the task is to check if there is a subset of the given array whose sum is equal to the given sum. Examples: Input: arr[] = [3, 34, 4, 12, 5, 2], sum = 9Output: TrueExplanation: There is a subset (4, 5) with sum 9. Input: arr[] = [3, 34,
      15+ min read

    • Introduction and Dynamic Programming solution to compute nCr%p
      Given three numbers n, r and p, compute value of nCr mod p. Example: Input: n = 10, r = 2, p = 13 Output: 6 Explanation: 10C2 is 45 and 45 % 13 is 6.We strongly recommend that you click here and practice it, before moving on to the solution.METHOD 1: (Using Dynamic Programming) A Simple Solution is
      15+ min read

    • Rod Cutting
      Given a rod of length n inches and an array price[]. price[i] denotes the value of a piece of length i. The task is to determine the maximum value obtainable by cutting up the rod and selling the pieces. Note: price[] is 1-indexed array. Input: price[] = [1, 5, 8, 9, 10, 17, 17, 20]Output: 22Explana
      15+ min read

    • Painting Fence Algorithm
      Given a fence with n posts and k colors, the task is to find out the number of ways of painting the fence so that not more than two consecutive posts have the same color. Examples: Input: n = 2, k = 4Output: 16Explanation: We have 4 colors and 2 posts.Ways when both posts have same color: 4 Ways whe
      15 min read

    • Longest Common Subsequence (LCS)
      Given two strings, s1 and s2, the task is to find the length of the Longest Common Subsequence. If there is no common subsequence, return 0. A subsequence is a string generated from the original string by deleting 0 or more characters, without changing the relative order of the remaining characters.
      15+ min read

    • Longest Increasing Subsequence (LIS)
      Given an array arr[] of size n, the task is to find the length of the Longest Increasing Subsequence (LIS) i.e., the longest possible subsequence in which the elements of the subsequence are sorted in increasing order. Examples: Input: arr[] = [3, 10, 2, 1, 20]Output: 3Explanation: The longest incre
      14 min read

    • Longest subsequence such that difference between adjacents is one
      Given an array arr[] of size n, the task is to find the longest subsequence such that the absolute difference between adjacent elements is 1. Examples: Input: arr[] = [10, 9, 4, 5, 4, 8, 6]Output: 3Explanation: The three possible subsequences of length 3 are [10, 9, 8], [4, 5, 4], and [4, 5, 6], whe
      15+ min read

    • Maximum size square sub-matrix with all 1s
      Given a binary matrix mat of size n * m, the task is to find out the maximum length of a side of a square sub-matrix with all 1s. Example: Input: mat = [ [0, 1, 1, 0, 1], [1, 1, 0, 1, 0], [0, 1, 1, 1, 0], [1, 1, 1, 1, 0], [1, 1, 1, 1, 1], [0, 0, 0, 0, 0] ] Output: 3Explanation: The maximum length of
      15+ min read

    • Min Cost Path
      You are given a 2D matrix cost[][] of dimensions m × n, where each cell represents the cost of traversing through that position. Your goal is to determine the minimum cost required to reach the bottom-right cell (m-1, n-1) starting from the top-left cell (0,0).The total cost of a path is the sum of
      15+ min read

    • Longest Common Substring (Space optimized DP solution)
      Given two strings ‘s1‘ and ‘s2‘, find the length of the longest common substring. Example: Input: s1 = “GeeksforGeeks”, s2 = “GeeksQuiz” Output : 5 Explanation:The longest common substring is “Geeks” and is of length 5. Input: s1 = “abcdxyz”, s2 = “xyzabcd” Output : 4Explanation:The longest common s
      7 min read

    • Count ways to reach the nth stair using step 1, 2 or 3
      A child is running up a staircase with n steps and can hop either 1 step, 2 steps, or 3 steps at a time. The task is to implement a method to count how many possible ways the child can run up the stairs. Examples: Input: 4Output: 7Explanation: There are seven ways: {1, 1, 1, 1}, {1, 2, 1}, {2, 1, 1}
      15+ min read

    • Grid Unique Paths - Count Paths in matrix
      Given an matrix of size m x n, the task is to find the count of all unique possible paths from top left to the bottom right with the constraints that from each cell we can either move only to the right or down. Examples: Input: m = 2, n = 2Output: 2Explanation: There are two paths(0, 0) -> (0, 1)
      15+ min read

    • Unique paths in a Grid with Obstacles
      Given a grid[][] of size m * n, let us assume we are starting at (1, 1) and our goal is to reach (m, n). At any instance, if we are on (x, y), we can either go to (x, y + 1) or (x + 1, y). The task is to find the number of unique paths if some obstacles are added to the grid.Note: An obstacle and sp
      15+ min read

    Medium problems on Dynamic programming

    • 0/1 Knapsack Problem
      Given n items where each item has some weight and profit associated with it and also given a bag with capacity W, [i.e., the bag can hold at most W weight in it]. The task is to put the items into the bag such that the sum of profits associated with them is the maximum possible. Note: The constraint
      15+ min read

    • Printing Items in 0/1 Knapsack
      Given weights and values of n items, put these items in a knapsack of capacity W to get the maximum total value in the knapsack. In other words, given two integer arrays, val[0..n-1] and wt[0..n-1] represent values and weights associated with n items respectively. Also given an integer W which repre
      12 min read

    • Unbounded Knapsack (Repetition of items allowed)
      Given a knapsack weight, say capacity and a set of n items with certain value vali and weight wti, The task is to fill the knapsack in such a way that we can get the maximum profit. This is different from the classical Knapsack problem, here we are allowed to use an unlimited number of instances of
      15+ min read

    • Egg Dropping Puzzle | DP-11
      You are given n identical eggs and you have access to a k-floored building from 1 to k. There exists a floor f where 0 <= f <= k such that any egg dropped from a floor higher than f will break, and any egg dropped from or below floor f will not break. There are a few rules given below: An egg
      15+ min read

    • Word Break
      Given a string s and y a dictionary of n words dictionary, check if s can be segmented into a sequence of valid words from the dictionary, separated by spaces. Examples: Input: s = "ilike", dictionary[] = ["i", "like", "gfg"]Output: trueExplanation: The string can be segmented as "i like". Input: s
      12 min read

    • Vertex Cover Problem (Dynamic Programming Solution for Tree)
      A vertex cover of an undirected graph is a subset of its vertices such that for every edge (u, v) of the graph, either ‘u’ or ‘v’ is in vertex cover. Although the name is Vertex Cover, the set covers all edges of the given graph. The problem to find minimum size vertex cover of a graph is NP complet
      15+ min read

    • Tile Stacking Problem
      Given integers n (the height of the tower), m (the maximum size of tiles available), and k (the maximum number of times each tile size can be used), the task is to calculate the number of distinct stable towers of height n that can be built. Note: A stable tower consists of exactly n tiles, each sta
      15+ min read

    • Box Stacking Problem
      Given three arrays height[], width[], and length[] of size n, where height[i], width[i], and length[i] represent the dimensions of a box. The task is to create a stack of boxes that is as tall as possible, but we can only stack a box on top of another box if the dimensions of the 2-D base of the low
      15+ min read

    • Partition a Set into Two Subsets of Equal Sum
      Given an array arr[], the task is to check if it can be partitioned into two parts such that the sum of elements in both parts is the same.Note: Each element is present in either the first subset or the second subset, but not in both. Examples: Input: arr[] = [1, 5, 11, 5]Output: true Explanation: T
      15+ min read

    • Travelling Salesman Problem using Dynamic Programming
      Given a 2d matrix cost[][] of size n where cost[i][j] denotes the cost of moving from city i to city j. The task is to complete a tour from city 0 (0-based index) to all other cities such that we visit each city exactly once and then at the end come back to city 0 at minimum cost. Note the differenc
      15 min read

    • Longest Palindromic Subsequence (LPS)
      Given a string s, find the length of the Longest Palindromic Subsequence in it. Note: The Longest Palindromic Subsequence (LPS) is the maximum-length subsequence of a given string that is also a Palindrome. Examples: Input: s = "bbabcbcab"Output: 7Explanation: Subsequence "babcbab" is the longest su
      15+ min read

    • Longest Common Increasing Subsequence (LCS + LIS)
      Given two arrays, a[] and b[], find the length of the longest common increasing subsequence(LCIS). LCIS refers to a subsequence that is present in both arrays and strictly increases.Prerequisites: LCS, LIS. Examples: Input: a[] = [3, 4, 9, 1], b[] = [5, 3, 8, 9, 10, 2, 1]Output: 2Explanation: The lo
      15+ min read

    • Find all distinct subset (or subsequence) sums of an array
      Given an array arr[] of size n, the task is to find a distinct sum that can be generated from the subsets of the given sets and return them in increasing order. It is given that the sum of array elements is small. Examples: Input: arr[] = [1, 2]Output: [0, 1, 2, 3]Explanation: Four distinct sums can
      15+ min read

    • Weighted Job Scheduling
      Given a 2D array jobs[][] of order n*3, where each element jobs[i] defines start time, end time, and the profit associated with the job. The task is to find the maximum profit you can take such that there are no two jobs with overlapping time ranges. Note: If the job ends at time X, it is allowed to
      15+ min read

    • Count Derangements (Permutation such that no element appears in its original position)
      A Derangement is a permutation of n elements, such that no element appears in its original position. For example, a derangement of [0, 1, 2, 3] is [2, 3, 1, 0].Given a number n, find the total number of Derangements of a set of n elements. Examples : Input: n = 2Output: 1Explanation: For two balls [
      12 min read

    • Minimum insertions to form a palindrome
      Given a string s, the task is to find the minimum number of characters to be inserted to convert it to a palindrome. Examples: Input: s = "geeks"Output: 3Explanation: "skgeegks" is a palindromic string, which requires 3 insertions. Input: s= "abcd"Output: 3Explanation: "abcdcba" is a palindromic str
      15+ min read

    • Ways to arrange Balls such that adjacent balls are of different types
      There are 'p' balls of type P, 'q' balls of type Q and 'r' balls of type R. Using the balls we want to create a straight line such that no two balls of the same type are adjacent.Examples : Input: p = 1, q = 1, r = 0Output: 2Explanation: There are only two arrangements PQ and QP Input: p = 1, q = 1,
      15+ min read

geeksforgeeks-footer-logo
Corporate & Communications Address:
A-143, 7th Floor, Sovereign Corporate Tower, Sector- 136, Noida, Uttar Pradesh (201305)
Registered Address:
K 061, Tower K, Gulshan Vivante Apartment, Sector 137, Noida, Gautam Buddh Nagar, Uttar Pradesh, 201305
GFG App on Play Store GFG App on App Store
Advertise with us
  • Company
  • About Us
  • Legal
  • Privacy Policy
  • In Media
  • Contact Us
  • Advertise with us
  • GFG Corporate Solution
  • Placement Training Program
  • Languages
  • Python
  • Java
  • C++
  • PHP
  • GoLang
  • SQL
  • R Language
  • Android Tutorial
  • Tutorials Archive
  • DSA
  • Data Structures
  • Algorithms
  • DSA for Beginners
  • Basic DSA Problems
  • DSA Roadmap
  • Top 100 DSA Interview Problems
  • DSA Roadmap by Sandeep Jain
  • All Cheat Sheets
  • Data Science & ML
  • Data Science With Python
  • Data Science For Beginner
  • Machine Learning
  • ML Maths
  • Data Visualisation
  • Pandas
  • NumPy
  • NLP
  • Deep Learning
  • Web Technologies
  • HTML
  • CSS
  • JavaScript
  • TypeScript
  • ReactJS
  • NextJS
  • Bootstrap
  • Web Design
  • Python Tutorial
  • Python Programming Examples
  • Python Projects
  • Python Tkinter
  • Python Web Scraping
  • OpenCV Tutorial
  • Python Interview Question
  • Django
  • Computer Science
  • Operating Systems
  • Computer Network
  • Database Management System
  • Software Engineering
  • Digital Logic Design
  • Engineering Maths
  • Software Development
  • Software Testing
  • DevOps
  • Git
  • Linux
  • AWS
  • Docker
  • Kubernetes
  • Azure
  • GCP
  • DevOps Roadmap
  • System Design
  • High Level Design
  • Low Level Design
  • UML Diagrams
  • Interview Guide
  • Design Patterns
  • OOAD
  • System Design Bootcamp
  • Interview Questions
  • Inteview Preparation
  • Competitive Programming
  • Top DS or Algo for CP
  • Company-Wise Recruitment Process
  • Company-Wise Preparation
  • Aptitude Preparation
  • Puzzles
  • School Subjects
  • Mathematics
  • Physics
  • Chemistry
  • Biology
  • Social Science
  • English Grammar
  • Commerce
  • World GK
  • GeeksforGeeks Videos
  • DSA
  • Python
  • Java
  • C++
  • Web Development
  • Data Science
  • CS Subjects
@GeeksforGeeks, Sanchhaya Education Private Limited, All rights reserved
We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy
Lightbox
Improvement
Suggest Changes
Help us improve. Share your suggestions to enhance the article. Contribute your expertise and make a difference in the GeeksforGeeks portal.
geeksforgeeks-suggest-icon
Create Improvement
Enhance the article with your expertise. Contribute to the GeeksforGeeks community and help create better learning resources for all.
geeksforgeeks-improvement-icon
Suggest Changes
min 4 words, max Words Limit:1000

Thank You!

Your suggestions are valuable to us.

What kind of Experience do you want to share?

Interview Experiences
Admission Experiences
Career Journeys
Work Experiences
Campus Experiences
Competitive Exam Experiences