Minimum partitions of String such that each part is at most K

Last Updated : 26 Sep, 2022

Given a string S of size N consisting of numerical digits 1-9 and a positive integer K, the task is to minimize the partitions of the string such that each part is at most K. If it’s impossible to partition the string print -1.

Examples:

Input: S = "3456", K = 45
Output: 2
Explanation: One possible way is to partition as 3, 45, 6.
Another possibility is 3, 4, 5, 6, which uses 3 partition.
No configuration needs less than 2 partitions. Hence, the answer is 2.

Input: S = "7891634", K = 21
Output: 5
Explanation: The minimum number of partitions is
7, 8, 9, 16, 3, 4, which uses 5 partitions.

Input: S = "67142849", K = 39
Output: 5

Approach: The approach to this problem is based on the below idea:

Make each partition have a value as large as possible with an upper limit of K.
An edge case is if it’s impossible to partition the string. It will happen if the maximum digit in the string is larger than K..

Follow the below steps to solve this problem:

Iterate over the characters of the string from i = 0 to N-1:
- If the number formed till now is at most K then keep on continuing with this partition. Otherwise, increase the number of partitions.
- Otherwise, If the current digit is larger than K, return -1.
The last number may not have been accounted for after iterating the string, so check if the last partition is greater than 0. If yes, then increment number of partitions (say ans) by 1.
Finally, return ans - 1, as we need to find the minimum number of partitions, which is one less than the numbers formed.

Below is the implementation of the above approach:

C++

// C++ program for above implementation  #include <bits/stdc++.h> using namespace std;  // Function to count the minimum number // of partitions int minimumCommas(string& s, int k) {     // Length of the string 's'.     int n = s.size();      // Store the current number formed.     long long cur = 0;      // 'ans' stores the final answer.     int ans = 0;      // Iterate over the string.     for (int i = 0; i < n; ++i) {          // If we can include this digit in the         // current number         if (cur * 10 + (s[i] - '0') <= k) {              // Include this digit in             // the current number.             cur = cur * 10 + (s[i] - '0');         }         else {              // If 'cur' is '0',             // it's impossible to partition             if (cur == 0 or cur > k) {                  // Return the integer '-1'                 return -1;             }             else {                  // Increment the answer 'ans'                 ans++;                  // Set cur to the current digit                 cur = s[i] - '0';             }         }     }      // If cur > 0, means the last number is cur     if (cur > 0 and cur <= k) {          // Increment the 'ans'         ans++;     }      // Return the number of partitions     return ans - 1; }  // Driver code int main() {     // Input     string S = "7891634";     int K = 21;      // Function call     cout << minimumCommas(S, K);     return 0; }

Java

// JAVA program for above implementation import java.util.*; class GFG {      // Function to count the minimum number     // of partitions     public static int minimumCommas(String s, int k)     {                // Length of the string 's'.         int n = s.length();          // Store the current number formed.         long cur = 0;          // 'ans' stores the final answer.         int ans = 0;          // Iterate over the string.         for (int i = 0; i < n; ++i) {             // If we can include this digit in the             // current number             if (cur * 10                     + Character.getNumericValue(s.charAt(i))                 <= k) {                  // Include this digit in                 // the current number.                 cur = cur * 10                       + Character.getNumericValue(                           s.charAt(i));             }             else {                  // If 'cur' is '0',                 // it's impossible to partition                 if (cur == 0 || cur > k) {                      // Return the integer '-1'                     return -1;                 }                 else {                      // Increment the answer 'ans'                     ans++;                      // Set cur to the current digit                     cur = Character.getNumericValue(                         s.charAt(i));                 }             }         }          // If cur > 0, means the last number is cur         if (cur > 0 && cur <= k) {              // Increment the 'ans'             ans++;         }          // Return the number of partitions         return ans - 1;     }      // Driver code     public static void main(String[] args)     {         // Input         String S = "7891634";         int K = 21;          // Function call         System.out.print(minimumCommas(S, K));     } }  // This code is contributed by Taranpreet

Python3

# Python code for the above approach  # Function to count the minimum number # of partitions def minimumCommas(s, k) :          # Length of the string 's'.     n = len(s)      # Store the current number formed.     cur = 0      # 'ans' stores the final answer.     ans = 0      # Iterate over the string.     for i in range(n) :          # If we can include this digit in the         # current number         if (cur * 10 + (ord(s[i]) - ord('0')) <= k) :              # Include this digit in             # the current number.             cur = cur * 10 + (ord(s[i]) - ord('0'))                  else :              # If 'cur' is '0',             # it's impossible to partition             if (cur == 0 or cur > k) :                  # Return the integer '-1'                 return -1                          else :                  # Increment the answer 'ans'                 ans += 1                  # Set cur to the current digit                 cur = (ord(s[i]) - ord('0'))                                 # If cur > 0, means the last number is cur     if (cur > 0 and cur <= k) :          # Increment the 'ans'         ans += 1           # Return the number of partitions     return ans - 1  # Driver code if __name__ == "__main__":          # Input     S = "7891634"     K = 21      # Function call     print(minimumCommas(S, K))          # This code is contributed by code_hunt.

// C# program for above implementation using System; class GFG {    // Function to count the minimum number   // of partitions   static int minimumCommas(string s, int k)   {      // Length of the string 's'.     int n = s.Length;      // Store the current number formed.     long cur = 0;      // 'ans' stores the final answer.     int ans = 0;      // Iterate over the string.     for (int i = 0; i < n; ++i) {       // If we can include this digit in the       // current number       if (cur * 10 + (s[i] - '0') <= k) {          // Include this digit in         // the current number.         cur = cur * 10 + (s[i] - '0');       }       else {          // If 'cur' is '0',         // it's impossible to partition         if (cur == 0 || cur > k) {            // Return the integer '-1'           return -1;         }         else {            // Increment the answer 'ans'           ans++;            // Set cur to the current digit           cur = s[i] - '0';         }       }     }      // If cur > 0, means the last number is cur     if (cur > 0 && cur <= k) {        // Increment the 'ans'       ans++;     }      // Return the number of partitions     return ans - 1;   }    // Driver code   public static void Main()   {     // Input     string S = "7891634";     int K = 21;      // Function call     Console.WriteLine(minimumCommas(S, K));   } }  // This code is contributed by Samim Hossain Mondal.

JavaScript

<script>  // JavaScript program for above implementation  // Function to count the minimum number // of partitions function minimumCommas(s, k) {      // Length of the string 's'.     let n = s.length;      // Store the current number formed.     let cur = 0;      // 'ans' stores the final answer.     let ans = 0;      // Iterate over the string.     for (let i = 0; i < n; ++i) {          // If we can include this digit in the         // current number         if (cur * 10 + (s.charCodeAt(i) - '0'.charCodeAt(0)) <= k) {              // Include this digit in             // the current number.             cur = cur * 10 + (s.charCodeAt(i) - '0'.charCodeAt(0));         }         else {              // If 'cur' is '0',             // it's impossible to partition             if (cur == 0 || cur > k) {                  // Return the integer '-1'                 return -1;             }             else {                  // Increment the answer 'ans'                 ans++;                  // Set cur to the current digit                 cur = s.charCodeAt(i) - '0'.charCodeAt(0);             }         }     }      // If cur > 0, means the last number is cur     if (cur > 0 && cur <= k) {          // Increment the 'ans'         ans++;     }      // Return the number of partitions     return ans - 1; }  // Driver code  // Input let S = "7891634"; let K = 21;  // Function call document.write(minimumCommas(S, K));  // This code is contributed // by shinjanpatra  </script>

Output

Time Complexity: O(N)
Auxiliary Space: O(1)

Minimize deletions such that sum of position of characters is at most K

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Minimum partitions of String such that each part is at most K

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