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.
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 <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 |
3
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
