# Longest equal substring with cost less than K

Given two string X and Y of the same length which consists of lowercase letters and also an integer K. The task is to find the maximum length up to which X can be changed to Y within the given cost K.
The cost of changing a character is given by the absolute difference between the ASCII value of the characters. That is, to change a character at index i, cost = |x[i] – Y[i]|

Examples:

Input: X = abcd, Y = bcde, K = 3
Output: 3
Explanation: A maximum of 3 characters can be changed because the cost to change each article is 1.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to maintain a prefix array of length N to store the absolute sum of the strings. That is, the cost to change the string X to Y. The following steps can be followed to compute the result:

• Maintain two pointers say i and j.
• In a while loop check if the difference between ith index and jth index of prefix array is greater than given cost or not.
• If the difference is greater than given cost then increase the j pointer to compensate for the cost else increase the i pointer.

Below is the implementation of the above approach:

## CPP

 // C++ program to find the // maximum length of equal substring // within a given cost #include using namespace std;    // Function to find the maximum length int solve(string X, string Y, int N, int K) {        int count[N + 1] = { 0 };     int sol = 0;     count[0] = 0;        // Fill the prefix array with     // the difference of letters     for (int i = 1; i <= N; i++) {            count[i] = count[i - 1] + abs(X[i - 1] - Y[i - 1]);     }        int j = 0;        for (int i = 1; i <= N; i++) {         while ((count[i] - count[j]) > K) {                j++;         }            // Update the maximum length         sol = max(sol, i - j);     }        return sol; }    // Driver code int main() {        int N = 4;        string X = "abcd", Y = "bcde";        int K = 3;        cout << solve(X, Y, N, K) << "\n";        return 0; }

## Java

 // Java program to find the // maximum length of equal subString // within a given cost class GFG {    // Function to find the maximum length static int solve(String X, String Y,                 int N, int K) {        int []count = new int[N + 1];     int sol = 0;     count[0] = 0;        // Fill the prefix array with     // the difference of letters     for (int i = 1; i <= N; i++)      {            count[i] = count[i - 1] +                  Math.abs(X.charAt(i - 1) -                  Y.charAt(i - 1));     }        int j = 0;        for (int i = 1; i <= N; i++)     {         while ((count[i] - count[j]) > K)         {             j++;         }            // Update the maximum length         sol = Math.max(sol, i - j);     }        return sol; }    // Driver code public static void main(String[] args) {        int N = 4;     String X = "abcd", Y = "bcde";     int K = 3;        System.out.print(solve(X, Y, N, K) + "\n"); } }    // This code is contributed by PrinciRaj1992

## Python3

 # Python3 program to find the # maximum length of equal subString # within a given cost    # Function to find the maximum length def solve(X, Y, N, K):     count = [0] * (N + 1);     sol = 0;     count[0] = 0;        # Fill the prefix array with     # the difference of letters     for i in range(1, N + 1):         count[i] = (count[i - 1] +                     abs(ord(X[i - 1]) -                      ord(Y[i - 1])));        j = 0;        for i in range(1, N + 1):         while ((count[i] - count[j]) > K):             j += 1;            # Update the maximum length         sol = max(sol, i - j);        return sol;    # Driver code if __name__ == '__main__':     N = 4;     X = "abcd";     Y = "bcde";     K = 3;        print(solve(X, Y, N, K));    # This code is contributed by PrinciRaj1992

## C#

 // C# program to find the  // maximum length of equal subString  // within a given cost  using System;    class GFG  {             // Function to find the maximum length      static int solve(string X, string Y,                      int N, int K)      {                 int []count = new int[N + 1];          int sol = 0;          count[0] = 0;                 // Fill the prefix array with          // the difference of letters          for (int i = 1; i <= N; i++)          {                     count[i] = count[i - 1] +                      Math.Abs(X[i - 1] -                      Y[i - 1]);          }                 int j = 0;                 for (int i = 1; i <= N; i++)          {              while ((count[i] - count[j]) > K)              {                  j++;              }                     // Update the maximum length              sol = Math.Max(sol, i - j);          }                 return sol;      }         // Driver code      public static void Main()      {                 int N = 4;          string X = "abcd", Y = "bcde";          int K = 3;                 Console.WriteLine(solve(X, Y, N, K) + "\n");      }  }     // This code is contributed by AnkitRai01

Output:

3

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