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Count number of binary strings without consecutive 1's
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Count number of binary strings without consecutive 1's

Last Updated : 16 Dec, 2024
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Given a positive integer n, the task is to count all possible distinct binary strings of length n such that there are no consecutive 1's.

Examples: 

Input: n = 3
Output: 5
Explanation: 5 strings are ("000", "001", "010", "100", "101").

Input: n = 2
Output: 3
Explanation: 3 strings are ("00", "01", "10").

Table of Content

  • Using Recursion - O(2^n) Time and O(n) Space
  • Using Top-Down DP (Memoization) - O(n) Time and O(n) Space
  • Using Bottom-Up DP (Tabulation) - O(n) Time and O(n) Space
  • Using Space Optimized DP - O(n) Time and O(1) Space
  • Using Matrix Exponentiation - O(logn) Time and O(logn) Space

Using Recursion - O(2^n) Time and O(n) Space

The idea is to explore two possible choices at each step of building the binary string. When constructing a string of length n, at each index, we have two choices: either place a 0 or place a 1. If we choose to place a 0, we can proceed to the next index. However, if we choose to place a 1, we must ensure that the next index does not contain a 1, so we skip the next index.

Mathematically the recurrence relation will look like the following:

countStrings(i) = countStrings(i+1) + countStrings(i+2)

Base Case:

  • countStrings(i) = 1, if i >= n.
C++ 
// C++ program to count number of binary // strings without consecutive 1's using recursion #include <bits/stdc++.h> using namespace std;  int countRecur(int i, int n) {          // Base case     if (i >= n) return 1;          // If we take 1 at ith index,      // we cannot have 1 at (i-1)     int take = countRecur(i+2, n);          // If we skip 1, we can consider     // 1 at i-1.     int noTake = countRecur(i+1, n);          return take + noTake; }  int countStrings(int n) {     return countRecur(0, n); }  int main() {     int n = 3;     cout << countStrings(n);      return 0; }
Java 
// Java program to count number of binary // strings without consecutive 1's using recursion  class GfG {     static int countRecur(int i, int n) {                // Base case         if (i >= n) return 1;          // If we take 1 at ith index,          // we cannot have 1 at (i-1)         int take = countRecur(i + 2, n);          // If we skip 1, we can consider         // 1 at i-1.         int noTake = countRecur(i + 1, n);          return take + noTake;     }      static int countStrings(int n) {         return countRecur(0, n);     }      public static void main(String[] args) {         int n = 3;         System.out.println(countStrings(n));     } }
Python 
# Python program to count number of binary # strings without consecutive 1's using recursion  def countRecur(i, n):        # Base case     if i >= n:         return 1      # If we take 1 at ith index,      # we cannot have 1 at (i-1)     take = countRecur(i + 2, n)      # If we skip 1, we can consider     # 1 at i-1.     noTake = countRecur(i + 1, n)      return take + noTake  def countStrings(n):     return countRecur(0, n)  if __name__ == "__main__":     n = 3     print(countStrings(n))
C# 
// C# program to count number of binary // strings without consecutive 1's using recursion  using System;  class GfG {     static int countRecur(int i, int n) {                // Base case         if (i >= n) return 1;          // If we take 1 at ith index,          // we cannot have 1 at (i-1)         int take = countRecur(i + 2, n);          // If we skip 1, we can consider         // 1 at i-1.         int noTake = countRecur(i + 1, n);          return take + noTake;     }      static int countStrings(int n) {         return countRecur(0, n);     }      static void Main(string[] args) {         int n = 3;         Console.WriteLine(countStrings(n));     } }
JavaScript 
// JavaScript program to count number of binary // strings without consecutive 1's using recursion  function countRecur(i, n) {      // Base case     if (i >= n) return 1;      // If we take 1 at ith index,      // we cannot have 1 at (i-1)     let take = countRecur(i + 2, n);      // If we skip 1, we can consider     // 1 at i-1.     let noTake = countRecur(i + 1, n);      return take + noTake; }  function countStrings(n) {     return countRecur(0, n); }  const n = 3; console.log(countStrings(n));

Output
5

Using Top-Down DP (Memoization) - O(n) Time and O(n) Space

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

1. Optimal Substructure: Number of ways to make binary string at i'th index depends on the optimal solutions of countStrings(i+1) and countStrings(i+2). By combining these substructures, we can efficiently calculate number of ways to make binary strings with consecutive 1's at index i.

2. Overlapping Subproblems: While applying a recursive approach in this problem, we notice that certain subproblems are computed multiple times. For example, while calculating countStrings(3), countStrings(4) and countStrings(5) are called. countStrings(4) will again call countStrings(5) which will lead to Overlapping Subproblems.

  • There is only one parameter: i that changes in the recursive solution. So we create a 1D array of size n for memoization.
  • We initialize this array as -1 to indicate nothing is computed initially.
  • Now we modify our recursive solution to first check if the value is -1, then only make recursive calls. This way, we avoid re-computations of the same subproblems.
C++ 
// C++ program to count number of binary // strings without consecutive 1's using memoization #include <bits/stdc++.h> using namespace std;  int countRecur(int i, int n, vector<int> &memo) {          // Base case     if (i >= n) return 1;          // If value is memoized     if (memo[i]!=-1) return memo[i];          // If we take 1 at ith index,      // we cannot have 1 at (i-1)     int take = countRecur(i+2, n, memo);          // If we skip 1, we can consider     // 1 at i-1.     int noTake = countRecur(i+1, n, memo);          return memo[i] = take + noTake; }  int countStrings(int n) {          vector<int> memo(n, -1);     return countRecur(0, n, memo); }  int main() {     int n = 3;     cout << countStrings(n);      return 0; }
Java 
// Java program to count number of binary // strings without consecutive 1's using memoization  class GfG {     static int countRecur(int i, int n, int[] memo) {                // Base case         if (i >= n) return 1;          // If value is memoized         if (memo[i] != -1) return memo[i];          // If we take 1 at ith index,          // we cannot have 1 at (i-1)         int take = countRecur(i + 2, n, memo);          // If we skip 1, we can consider         // 1 at i-1.         int noTake = countRecur(i + 1, n, memo);          return memo[i] = take + noTake;     }      static int countStrings(int n) {         int[] memo = new int[n];         for (int j = 0; j < n; j++) memo[j] = -1;         return countRecur(0, n, memo);     }      public static void main(String[] args) {         int n = 3;         System.out.println(countStrings(n));     } }
Python 
# Python program to count number of binary # strings without consecutive 1's using memoization  def countRecur(i, n, memo):        # Base case     if i >= n:         return 1      # If value is memoized     if memo[i] != -1:         return memo[i]      # If we take 1 at ith index,      # we cannot have 1 at (i-1)     take = countRecur(i + 2, n, memo)      # If we skip 1, we can consider     # 1 at i-1.     noTake = countRecur(i + 1, n, memo)      memo[i] = take + noTake     return memo[i]  def countStrings(n):     memo = [-1] * n     return countRecur(0, n, memo)  if __name__ == "__main__":     n = 3     print(countStrings(n))
C# 
// C# program to count number of binary // strings without consecutive 1's using memoization  using System;  class GfG {     static int countRecur(int i, int n, int[] memo) {                // Base case         if (i >= n) return 1;          // If value is memoized         if (memo[i] != -1) return memo[i];          // If we take 1 at ith index,          // we cannot have 1 at (i-1)         int take = countRecur(i + 2, n, memo);          // If we skip 1, we can consider         // 1 at i-1.         int noTake = countRecur(i + 1, n, memo);          return memo[i] = take + noTake;     }      static int countStrings(int n) {         int[] memo = new int[n];         for (int j = 0; j < n; j++) memo[j] = -1;         return countRecur(0, n, memo);     }      static void Main(string[] args) {         int n = 3;         Console.WriteLine(countStrings(n));     } }
JavaScript 
// JavaScript program to count number of binary // strings without consecutive 1's using memoization  function countRecur(i, n, memo) {      // Base case     if (i >= n) return 1;      // If value is memoized     if (memo[i] !== -1) return memo[i];      // If we take 1 at ith index,      // we cannot have 1 at (i-1)     let take = countRecur(i + 2, n, memo);      // If we skip 1, we can consider     // 1 at i-1.     let noTake = countRecur(i + 1, n, memo);      return memo[i] = take + noTake; }  function countStrings(n) {     let memo = Array(n).fill(-1);     return countRecur(0, n, memo); }  const n = 3; console.log(countStrings(n));

Output
5

Using Bottom-Up DP (Tabulation) - O(n) Time and O(n) Space

The idea is to fill the DP table based on next values. For each index, we can either place 1 or 0. The array is filled in an iterative manner from i = n-1 to i = 0.

The dynamic programming relation is as follows: 

  • dp[i] = dp[i+1] + dp[i+2]
C++ 
// C++ program to count number of binary // strings without consecutive 1's using tabulation #include <bits/stdc++.h> using namespace std;  int countStrings(int n) {     if (n == 1)         return 2;     if (n == 2)         return 3;      vector<int> dp(n);     dp[n - 1] = 2;     dp[n - 2] = 3;      for (int i = n - 3; i >= 0; i--) {         dp[i] = dp[i + 1] + dp[i + 2];     }      return dp[0]; }  int main() {        int n = 3;     cout << countStrings(n);      return 0; }
Java 
// Java program to count number of binary // strings without consecutive 1's using tabulation  class GfG {     static int countStrings(int n) {         if (n == 1) return 2;         if (n == 2) return 3;          int[] dp = new int[n];         dp[n - 1] = 2;         dp[n - 2] = 3;          for (int i = n - 3; i >= 0; i--) {             dp[i] = dp[i + 1] + dp[i + 2];         }          return dp[0];     }      public static void main(String[] args) {         int n = 3;         System.out.println(countStrings(n));     } }
Python 
# Python program to count number of binary # strings without consecutive 1's using tabulation  def countStrings(n):     if n == 1:         return 2     if n == 2:         return 3      dp = [0] * n     dp[n - 1] = 2     dp[n - 2] = 3      for i in range(n - 3, -1, -1):         dp[i] = dp[i + 1] + dp[i + 2]      return dp[0]  if __name__ == "__main__":     n = 3     print(countStrings(n))
C# 
// C# program to count number of binary // strings without consecutive 1's using tabulation  using System;  class GfG {     static int countStrings(int n) {         if (n == 1) return 2;         if (n == 2) return 3;          int[] dp = new int[n];         dp[n - 1] = 2;         dp[n - 2] = 3;          for (int i = n - 3; i >= 0; i--) {             dp[i] = dp[i + 1] + dp[i + 2];         }          return dp[0];     }      static void Main(string[] args) {         int n = 3;         Console.WriteLine(countStrings(n));     } }
JavaScript 
// JavaScript program to count number of binary // strings without consecutive 1's using tabulation  function countStrings(n) {     if (n === 1) return 2;     if (n === 2) return 3;      const dp = new Array(n).fill(0);     dp[n - 1] = 2;     dp[n - 2] = 3;      for (let i = n - 3; i >= 0; i--) {         dp[i] = dp[i + 1] + dp[i + 2];     }      return dp[0]; }  const n = 3; console.log(countStrings(n));

Output
5

Using Space Optimized DP - O(n) Time and O(1) Space

In previous approach of dynamic programming we have derive the relation between states as given below:

  • dp[i] = dp[i+1] + dp[i+2]

We observe that for calculating dp[i] state we only need dp[i+1] and dp[i+2]. There is no need to store all the next states.

C++ 
// C++ program to count number of binary // strings without consecutive 1's // using space optimised dp #include <bits/stdc++.h> using namespace std;  int countStrings(int n) {     if (n == 1)         return 2;     if (n == 2)         return 3;      int prev1 = 3, prev2 = 2;      for (int i = n - 3; i >= 0; i--) {         int curr = prev1 + prev2;          prev2 = prev1;         prev1 = curr;     }      return prev1; }  int main() {     int n = 3;     cout << countStrings(n);      return 0; }
Java 
// Java program to count number of binary // strings without consecutive 1's // using space optimised dp  class GfG {     static int countStrings(int n) {         if (n == 1)             return 2;         if (n == 2)             return 3;          int prev1 = 3, prev2 = 2;          for (int i = n - 3; i >= 0; i--) {             int curr = prev1 + prev2;              prev2 = prev1;             prev1 = curr;         }          return prev1;     }      public static void main(String[] args) {         int n = 3;         System.out.println(countStrings(n));     } }
Python 
# Python program to count number of binary # strings without consecutive 1's # using space optimised dp  def countStrings(n):     if n == 1:         return 2     if n == 2:         return 3      prev1, prev2 = 3, 2      for i in range(n - 3, -1, -1):         curr = prev1 + prev2          prev2 = prev1         prev1 = curr      return prev1  if __name__ == "__main__":     n = 3     print(countStrings(n))
C# 
// C# program to count number of binary // strings without consecutive 1's // using space optimised dp  using System;  class GfG {     static int countStrings(int n) {         if (n == 1)             return 2;         if (n == 2)             return 3;          int prev1 = 3, prev2 = 2;          for (int i = n - 3; i >= 0; i--) {             int curr = prev1 + prev2;              prev2 = prev1;             prev1 = curr;         }          return prev1;     }      static void Main(string[] args) {         int n = 3;         Console.WriteLine(countStrings(n));     } }
JavaScript 
// JavaScript program to count number of binary // strings without consecutive 1's // using space optimised dp  function countStrings(n) {     if (n === 1)         return 2;     if (n === 2)         return 3;      let prev1 = 3, prev2 = 2;      for (let i = n - 3; i >= 0; i--) {         let curr = prev1 + prev2;          prev2 = prev1;         prev1 = curr;     }      return prev1; }  const n = 3; console.log(countStrings(n));

Output
5

Using Matrix Exponentiation - O(logn) Time and O(logn) Space

The recurrence relation to count binary strings of length i is:

  • F(i) = F(i-1) + F(i-2)

with bases cases F(1) = 2 and F(2) = 3.

Matrix exponentiation can be used to solve this problem in O(logn) time. Refer to Matrix Exponentiation for detailed approach.

C++ 
// C++ program to count number of binary // strings without consecutive 1's #include <bits/stdc++.h> using namespace std;  // Function to mutiply two matrices vector<vector<int>> multiply(vector<vector<int>>& v1, vector<vector<int>>& v2) {     vector<vector<int>> ans(2, vector<int>(2, 0));      for (int i = 0; i < 2; i++) {         for (int j = 0; j < 2; j++) {             for (int k = 0; k < 2; k++) {                 ans[i][j] += v1[i][k] * v2[k][j];             }         }     }      return ans; }  // Function to find matrix to the power n. vector<vector<int>> power(vector<vector<int>> &v, int n) {     if (n == 0) {         return {{1, 0}, {0, 1}};     }      vector<vector<int>> res = power(v, n / 2);      res = multiply(res, res);      if (n % 2 == 1)         res = multiply(res, v);      return res; }  int countStrings(int n) {     if (n == 2)         return 3;     if (n == 1)         return 2;     vector<vector<int>> v = {{1, 1}, {1, 0}};     vector<vector<int>> p = power(v, n - 2);      int ans = p[0][0] * 3 + p[0][1] * 2;     return ans; }  int main() {     int n = 3;     cout << countStrings(n);      return 0; }
Java 
// Java program to count number of binary // strings without consecutive 1's  class GfG {      // Function to multiply two matrices     static int[][] multiply(int[][] v1, int[][] v2) {         int[][] ans = new int[2][2];          for (int i = 0; i < 2; i++) {             for (int j = 0; j < 2; j++) {                 for (int k = 0; k < 2; k++) {                     ans[i][j] += v1[i][k] * v2[k][j];                 }             }         }          return ans;     }      // Function to find matrix to the power n     static int[][] power(int[][] v, int n) {         if (n == 0) {             return new int[][] { { 1, 0 }, { 0, 1 } };         }          int[][] res = power(v, n / 2);         res = multiply(res, res);          if (n % 2 == 1) {             res = multiply(res, v);         }          return res;     }      static int countStrings(int n) {         if (n == 2)             return 3;         if (n == 1)             return 2;          int[][] v = { { 1, 1 }, { 1, 0 } };         int[][] p = power(v, n - 2);          int ans = p[0][0] * 3 + p[0][1] * 2;         return ans;     }      public static void main(String[] args) {         int n = 3;         System.out.println(countStrings(n));     } }
Python 
# Python program to count number of binary # strings without consecutive 1's  # Function to multiply two matrices def multiply(v1, v2):     ans = [[0, 0], [0, 0]]      for i in range(2):         for j in range(2):             for k in range(2):                 ans[i][j] += v1[i][k] * v2[k][j]      return ans  # Function to find matrix to the power n def power(v, n):     if n == 0:         return [[1, 0], [0, 1]]      res = power(v, n // 2)     res = multiply(res, res)      if n % 2 == 1:         res = multiply(res, v)      return res  def countStrings(n):     if n == 2:         return 3     if n == 1:         return 2      v = [[1, 1], [1, 0]]     p = power(v, n - 2)      ans = p[0][0] * 3 + p[0][1] * 2     return ans  if __name__ == "__main__":     n = 3     print(countStrings(n))
C# 
// C# program to count number of binary // strings without consecutive 1's  using System;  class GfG {      // Function to multiply two matrices     static int[, ] multiply(int[, ] v1, int[, ] v2) {         int[, ] ans = new int[2, 2];          for (int i = 0; i < 2; i++) {             for (int j = 0; j < 2; j++) {                 for (int k = 0; k < 2; k++) {                     ans[i, j] += v1[i, k] * v2[k, j];                 }             }         }          return ans;     }      // Function to find matrix to the power n     static int[, ] power(int[, ] v, int n) {         if (n == 0) {             return new int[, ] { { 1, 0 }, { 0, 1 } };         }          int[, ] res = power(v, n / 2);         res = multiply(res, res);          if (n % 2 == 1) {             res = multiply(res, v);         }          return res;     }      static int countStrings(int n) {         if (n == 2)             return 3;         if (n == 1)             return 2;          int[, ] v = { { 1, 1 }, { 1, 0 } };         int[, ] p = power(v, n - 2);          int ans = p[0, 0] * 3 + p[0, 1] * 2;         return ans;     }      static void Main(string[] args) {         int n = 3;         Console.WriteLine(countStrings(n));     } }
JavaScript 
// JavaScript program to count number of binary // strings without consecutive 1's  // Function to multiply two matrices function multiply(v1, v2) {     let ans = [ [ 0, 0 ], [ 0, 0 ] ];      for (let i = 0; i < 2; i++) {         for (let j = 0; j < 2; j++) {             for (let k = 0; k < 2; k++) {                 ans[i][j] += v1[i][k] * v2[k][j];             }         }     }      return ans; }  // Function to find matrix to the power n function power(v, n) {     if (n === 0) {         return [ [ 1, 0 ], [ 0, 1 ] ];     }      let res = power(v, Math.floor(n / 2));     res = multiply(res, res);      if (n % 2 === 1) {         res = multiply(res, v);     }      return res; }  function countStrings(n) {     if (n === 2)         return 3;     if (n === 1)         return 2;      let v = [ [ 1, 1 ], [ 1, 0 ] ];     let p = power(v, n - 2);      let ans = p[0][0] * 3 + p[0][1] * 2;     return ans; }  const n = 3; console.log(countStrings(n));

Output
5

Related article: 

  • 1 to n bit numbers with no consecutive 1s in binary representation.

Next Article
Count number of binary strings without consecutive 1's

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    Given a string s, the task is to reverse the string. Reversing a string means rearranging the characters such that the first character becomes the last, the second character becomes second last and so on.Examples:Input: s = "GeeksforGeeks"Output: "skeeGrofskeeG"Explanation : The first character G mo
    13 min read
    Left Rotation of a String
    Given a string s and an integer d, the task is to left rotate the string by d positions.Examples:Input: s = "GeeksforGeeks", d = 2Output: "eksforGeeksGe" Explanation: After the first rotation, string s becomes "eeksforGeeksG" and after the second rotation, it becomes "eksforGeeksGe".Input: s = "qwer
    15+ min read
    Sort string of characters
    Given a string of lowercase characters from 'a' - 'z'. We need to write a program to print the characters of this string in sorted order.Examples: Input : "dcab" Output : "abcd"Input : "geeksforgeeks"Output : "eeeefggkkorss"Naive Approach - O(n Log n) TimeA simple approach is to use sorting algorith
    5 min read
    Frequency of Characters in Alphabetical Order
    Given a string s, the task is to print the frequency of each of the characters of s in alphabetical order.Example: Input: s = "aabccccddd" Output: a2b1c4d3 Since it is already in alphabetical order, the frequency of the characters is returned for each character. Input: s = "geeksforgeeks" Output: e4
    9 min read
    Swap characters in a String
    Given a String S of length N, two integers B and C, the task is to traverse characters starting from the beginning, swapping a character with the character after C places from it, i.e. swap characters at position i and (i + C)%N. Repeat this process B times, advancing one position at a time. Your ta
    14 min read
    C Program to Find the Length of a String
    The length of a string is the number of characters in it without including the null character (‘\0’). In this article, we will learn how to find the length of a string in C.The easiest way to find the string length is by using strlen() function from the C strings library. Let's take a look at an exa
    2 min read
    How to insert characters in a string at a certain position?
    Given a string str and an array of indices chars[] that describes the indices in the original string where the characters will be added. For this post, let the character to be inserted in star (*). Each star should be inserted before the character at the given index. Return the modified string after
    7 min read
    Check if two strings are same or not
    Given two strings, the task is to check if these two strings are identical(same) or not. Consider case sensitivity.Examples:Input: s1 = "abc", s2 = "abc" Output: Yes Input: s1 = "", s2 = "" Output: Yes Input: s1 = "GeeksforGeeks", s2 = "Geeks" Output: No Approach - By Using (==) in C++/Python/C#, eq
    7 min read
    Concatenating Two Strings in C
    Concatenating two strings means appending one string at the end of another string. In this article, we will learn how to concatenate two strings in C.The most straightforward method to concatenate two strings is by using strcat() function. Let's take a look at an example:C#include <stdio.h> #i
    2 min read
    Remove all occurrences of a character in a string
    Given a string and a character, remove all the occurrences of the character in the string.Examples: Input : s = "geeksforgeeks" c = 'e'Output : s = "gksforgks"Input : s = "geeksforgeeks" c = 'g'Output : s = "eeksforeeks"Input : s = "geeksforgeeks" c = 'k'Output : s = "geesforgees"Using Built-In Meth
    2 min read

    Binary String

    Check if all bits can be made same by single flip
    Given a binary string, find if it is possible to make all its digits equal (either all 0's or all 1's) by flipping exactly one bit. Input: 101Output: YeExplanation: In 101, the 0 can be flipped to make it all 1Input: 11Output: NoExplanation: No matter whichever digit you flip, you will not get the d
    5 min read
    Number of flips to make binary string alternate | Set 1
    Given a binary string, that is it contains only 0s and 1s. We need to make this string a sequence of alternate characters by flipping some of the bits, our goal is to minimize the number of bits to be flipped. Examples : Input : str = “001” Output : 1 Minimum number of flips required = 1 We can flip
    8 min read
    Binary representation of next number
    Given a binary string that represents binary representation of positive number n, the task is to find the binary representation of n+1. The binary input may or may not fit in an integer, so we need to return a string.Examples: Input: s = "10011"Output: "10100"Explanation: Here n = (19)10 = (10011)2n
    6 min read
    Min flips of continuous characters to make all characters same in a string
    Given a string consisting only of 1's and 0's. In one flip we can change any continuous sequence of this string. Find this minimum number of flips so the string consist of same characters only.Examples: Input : 00011110001110Output : 2We need to convert 1's sequenceso string consist of all 0's.Input
    8 min read
    Generate all binary strings without consecutive 1's
    Given an integer n, the task is to generate all binary strings of size n without consecutive 1's.Examples: Input : n = 4Output : 0000 0001 0010 0100 0101 1000 1001 1010Input : n = 3Output : 000 001 010 100 101Approach:The idea is to generate all binary strings of length n without consecutive 1's usi
    6 min read
    K'th bit in a binary representation with n iterations
    Given a decimal number m. Consider its binary representation string and apply n iterations. In each iteration, replace the character 0 with the string 01, and 1 with 10. Find the kth (1-based indexing) character in the string after the nth iterationExamples: Input: m = 5, n = 2, k = 5Output: 0Explan
    15+ min read

    Substring and Subsequence

    All substrings of a given String
    Given a string s, containing lowercase alphabetical characters. The task is to print all non-empty substrings of the given string.Examples : Input : s = "abc"Output : "a", "ab", "abc", "b", "bc", "c"Input : s = "ab"Output : "a", "ab", "b"Input : s = "a"Output : "a"[Expected Approach] - Using Iterati
    8 min read
    Print all subsequences of a string
    Given a string, we have to find out all its subsequences of it. A String is said to be a subsequence of another String, if it can be obtained by deleting 0 or more character without changing its order.Examples: Input : abOutput : "", "a", "b", "ab"Input : abcOutput : "", "a", "b", "c", "ab", "ac", "
    12 min read
    Count Distinct Subsequences
    Given a string str of length n, your task is to find the count of distinct subsequences of it.Examples: Input: str = "gfg"Output: 7Explanation: The seven distinct subsequences are "", "g", "f", "gf", "fg", "gg" and "gfg" Input: str = "ggg"Output: 4Explanation: The four distinct subsequences are "",
    13 min read
    Count distinct occurrences as a subsequence
    Given two strings pat and txt, where pat is always shorter than txt, count the distinct occurrences of pat as a subsequence in txt.Examples: Input: txt = abba, pat = abaOutput: 2Explanation: pat appears in txt as below three subsequences.[abba], [abba]Input: txt = banana, pat = banOutput: 3Explanati
    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
    Shortest Superstring Problem
    Given a set of n strings arr[], find the smallest string that contains each string in the given set as substring. We may assume that no string in arr[] is substring of another string.Examples: Input: arr[] = {"geeks", "quiz", "for"}Output: geeksquizforExplanation: "geeksquizfor" contains all the thr
    15+ min read
    Printing Shortest Common Supersequence
    Given two strings s1 and s2, find the shortest string which has both s1 and s2 as its sub-sequences. If multiple shortest super-sequence exists, print any one of them.Examples:Input: s1 = "geek", s2 = "eke"Output: geekeExplanation: String "geeke" has both string "geek" and "eke" as subsequences.Inpu
    9 min read
    Shortest Common Supersequence
    Given two strings s1 and s2, the task is to find the length of the shortest string that has both s1 and s2 as subsequences.Examples: Input: s1 = "geek", s2 = "eke"Output: 5Explanation: String "geeke" has both string "geek" and "eke" as subsequences.Input: s1 = "AGGTAB", s2 = "GXTXAYB"Output: 9Explan
    15+ min read
    Longest Repeating Subsequence
    Given a string s, the task is to find the length of the longest repeating subsequence, such that the two subsequences don't have the same string character at the same position, i.e. any ith character in the two subsequences shouldn't have the same index in the original string. Examples:Input: s= "ab
    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. Longest Palindromic SubsequenceExamples:Input: s = "bbabcbcab"Output: 7Explanation: Subsequen
    15+ min read
    Longest Palindromic Substring
    Given a string s, the task is to find the longest substring which is a palindrome. If there are multiple answers, then return the first appearing substring.Examples:Input: s = "forgeeksskeegfor" Output: "geeksskeeg"Explanation: There are several possible palindromic substrings like "kssk", "ss", "ee
    12 min read

    Palindrome

    C Program to Check for Palindrome String
    A string is said to be palindrome if the reverse of the string is the same as the string. In this article, we will learn how to check whether the given string is palindrome or not using C program.The simplest method to check for palindrome string is to reverse the given string and store it in a temp
    4 min read
    Check if a given string is a rotation of a palindrome
    Given a string, check if it is a rotation of a palindrome. For example your function should return true for "aab" as it is a rotation of "aba". Examples: Input: str = "aaaad" Output: 1 // "aaaad" is a rotation of a palindrome "aadaa" Input: str = "abcd" Output: 0 // "abcd" is not a rotation of any p
    15+ min read
    Check if characters of a given string can be rearranged to form a palindrome
    Given a string, Check if the characters of the given string can be rearranged to form a palindrome. For example characters of "geeksogeeks" can be rearranged to form a palindrome "geeksoskeeg", but characters of "geeksforgeeks" cannot be rearranged to form a palindrome. Recommended PracticeAnagram P
    14 min read
    Online algorithm for checking palindrome in a stream
    Given a stream of characters (characters are received one by one), write a function that prints 'Yes' if a character makes the complete string palindrome, else prints 'No'. Examples:Input: str[] = "abcba"Output: a Yes // "a" is palindrome b No // "ab" is not palindrome c No // "abc" is not palindrom
    15+ min read
    Print all Palindromic Partitions of a String using Bit Manipulation
    Given a string, find all possible palindromic partitions of a given string. Note that this problem is different from Palindrome Partitioning Problem, there the task was to find the partitioning with minimum cuts in input string. Here we need to print all possible partitions. Example: Input: nitinOut
    10 min read
    Minimum Characters to Add at Front for Palindrome
    Given a string s, the task is to find the minimum number of characters to be added to the front of s to make it palindrome. A palindrome string is a sequence of characters that reads the same forward and backward. Examples: Input: s = "abc"Output: 2Explanation: We can make above string palindrome as
    12 min read
    Make largest palindrome by changing at most K-digits
    You are given a string s consisting of digits (0-9) and an integer k. Convert the string into a palindrome by changing at most k digits. If multiple palindromes are possible, return the lexicographically largest one. If it's impossible to form a palindrome with k changes, return "Not Possible".Examp
    14 min read
    Minimum Deletions to Make a String Palindrome
    Given a string s of length n, the task is to remove or delete the minimum number of characters from the string so that the resultant string is a palindrome. Note: The order of characters should be maintained. Examples : Input : s = "aebcbda"Output : 2Explanation: Remove characters 'e' and 'd'. Resul
    15+ min read
    Minimum insertions to form a palindrome with permutations allowed
    Given a string of lowercase letters. Find minimum characters to be inserted in the string so that it can become palindrome. We can change the positions of characters in the string.Examples: Input: geeksforgeeksOutput: 2Explanation: geeksforgeeks can be changed as: geeksroforskeeg or geeksorfroskeeg
    5 min read
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