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 i^th index and j^th 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 <bits/stdc++.h> 
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:

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My Personal Notes

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

CPP

Java

Python3

C#

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