**Maximizing Equal Substrings Within a Given Cost Using Sliding Window Technique**
\n# Intuition\nWhen given two strings and a cost constraint, the problem essentially boils down to finding the longest substring that can be transformed within 
\n# Intuition\nWhen given two strings and a cost constraint, the problem essentially boils down to finding the longest substring that can be transformed within the given cost. My first thought was to use a sliding window approach. This method is efficient for problems involving subarrays or substrings and constraints, as it dynamically adjusts the window size to stay within the allowed cost.\n\n# Approach\nThe key idea is to maintain a window that keeps track of the total transformation cost. Here\'s a step-by-step breakdown:\n\n1. **Initialization**: We initialize several variables:\n   - `n` to store the length of the strings.\n   - `res` to store the maximum length of the valid substring.\n   - `win` to store the current total cost within the window.\n   - `l` as the left pointer of our sliding window.\n\n2. **Iterate with the Right Pointer**: We iterate through the string using a right pointer `r`.\n   - For each character, we calculate the transformation cost to change `s[r]` to `t[r]` and add it to `win`.\n\n3. **Adjust the Window**: If the total cost `win` exceeds `maxCost`, we need to shrink the window from the left side to bring the cost back within the limit:\n   - We do this by incrementing the left pointer `l` and subtracting the transformation cost of the character at the left pointer from `win`.\n\n4. **Update the Result**: After adjusting the window, we update `res` with the maximum length of the valid window (`r - l + 1`).\n\n5. **Return the Result**: Finally, we return the maximum length stored in `res`.\n\n\n\n# Complexity\n- **Time complexity**: The time complexity is $$O(n)$$ because we traverse the string once with the right pointer, and the left pointer also moves at most `n` times.\n- **Space complexity**: The space complexity is $$O(1)$$ as we use a constant amount of extra space regardless of the input size.\n\n**#SlidingWindow #Algorithm #StringManipulation #ComplexityAnalysis**\n# Code\n```\nclass Solution {\n    public int equalSubstring(String s, String t, int maxCost) {\n        int n = s.length();  // length of the strings\n        int res = 0;         // result to store the maximum length\n        int win = 0;         // variable to store the current total cost\n        int l = 0;           // left pointer of the sliding window\n\n        // Iterate through the string with the right pointer r\n        for(int r = 0; r < n; r++) {\n            // Add the transformation cost for the current character\n            win += Math.abs(s.charAt(r) - t.charAt(r));\n            \n            // If the current cost exceeds maxCost, adjust the left pointer\n            while(win > maxCost) {\n                win -= Math.abs(s.charAt(l) - t.charAt(l));\n                l++;\n            }\n            \n            // Update the result with the maximum length found\n            res = Math.max(res, r - l + 1);\n        }\n        return res;  // Return the result\n    }\n}\n```
['Java']