LeetCode 1208: Get Equal Substrings Within Budget Solution

Master LeetCode problem 1208 (Get Equal Substrings Within Budget), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1208. Get Equal Substrings Within Budget

Medium

String Binary Search Sliding Window Prefix Sum

Problem Explanation

Explanation:

To solve this problem, we can use a sliding window approach. We will maintain a window that represents the substring of s that we are considering. We will keep track of the cost of changing each character within the window and slide the window to the right while keeping the total cost within the given maxCost.

Here are the steps:

Initialize variables: start and end to represent the window, currCost to track the current cost within the window, and maxLength to store the length of the maximum substring.
Iterate over the characters of s using the sliding window approach:
- Update currCost by adding the cost of changing the character at end.
- If currCost exceeds maxCost, move the start of the window to the right and subtract the cost of changing the character at start from currCost.
- Update maxLength with the maximum length of the substring with cost less than or equal to maxCost.
Return maxLength.

The time complexity of this algorithm is O(n) where n is the length of the input string s. The space complexity is O(1) as we are using a constant amount of extra space.

Solution Code

class Solution {
    public int equalSubstring(String s, String t, int maxCost) {
        int start = 0, end = 0, currCost = 0, maxLength = 0;
        
        for (; end < s.length(); end++) {
            currCost += Math.abs(s.charAt(end) - t.charAt(end));
            
            while (currCost > maxCost) {
                currCost -= Math.abs(s.charAt(start) - t.charAt(start));
                start++;
            }
            
            maxLength = Math.max(maxLength, end - start + 1);
        }
        
        return maxLength;
    }
}

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