Must Watch✅💯Easiest Sliding Window Approach With Detailed Explanation💯✅
Intuition\n- Given two strings s and t of equal length, and an integer maxCost, the task is to find the maximum length of a substring that can be made equal to 
# Intuition\n- Given two strings s and t of equal length, and an integer maxCost, the task is to find the maximum length of a substring that can be made equal to the corresponding substring of t by changing characters in s, where the cost of changing a character is the absolute difference between their ASCII values. The total cost of changes must not exceed maxCost.\n\n# Approach\n1. Sliding Window Technique:\n\n    - We use a sliding window to keep track of the current substring being compared between s and t.\nThe idea is to maintain a window of characters from s and check the cost of converting this substring into the corresponding characters in t.\n\n    - We expand the window by adding characters from s and calculating the transformation cost between s and t. If the cost exceeds maxCost, we shrink the window from the left.\n\n2. Window Adjustment:\n\n    - For each character at index j in s, we calculate the difference with t[j] and add it to the current window cost.\nIf the window cost exceeds maxCost, we shrink the window from the left by incrementing the starting index i and reducing the cost accordingly.\n\n3. Max Length Update:\n\n    - While maintaining the sliding window, we keep track of the maximum length of the substring that satisfies the condition (where the total transformation cost does not exceed maxCost).\n\n# Complexity\n- Time complexity:\nO(n) where n is the length of the strings s and t. Each character is processed once as part of the sliding window, resulting in linear time complexity.\n\n- Space complexity:\nO(1) as the algorithm uses only a few extra variables (for tracking the indices and cost), and the space used does not depend on the size of the input strings.\n# Code\n```python3 []\nclass Solution:\n    def equalSubstring(self, s: str, t: str, maxCost: int) -> int:\n        i = 0\n        curr_wind = 0  # Initialize window cost\n        max_length = 0  # Store maximum valid substring length\n        \n        for j in range(len(s)):  # Iterate over each character in the string\n            # Add cost of current character difference to window\n            curr_wind += abs(ord(s[j]) - ord(t[j]))\n            \n            # If window cost exceeds maxCost, shrink the window\n            while curr_wind > maxCost:\n                curr_wind -= abs(ord(s[i]) - ord(t[i]))  # Remove cost of leftmost character\n                i += 1  # Move the start of the window to the right\n            \n            # Update maximum length of valid window\n            max_length = max(max_length, j - i + 1)\n        \n        return max_length           \n```\n```Java []\nclass Solution {\n    public int equalSubstring(String s, String t, int maxCost) {\n        int i = 0, currWind = 0, maxLength = 0;\n        \n        for (int j = 0; j < s.length(); j++) {\n            // Add the cost of transforming s[j] to t[j] to the current window\n            currWind += Math.abs(s.charAt(j) - t.charAt(j));\n            \n            // If the current window cost exceeds maxCost, shrink the window from the left\n            while (currWind > maxCost) {\n                currWind -= Math.abs(s.charAt(i) - t.charAt(i));\n                i++;\n            }\n            \n            // Update the maximum length of a valid substring\n            maxLength = Math.max(maxLength, j - i + 1);\n        }\n        \n        return maxLength;\n    }\n}\n       \n```\n```C++ []\nclass Solution {\npublic:\n    int equalSubstring(string s, string t, int maxCost) {\n        int i = 0, currWind = 0, maxLength = 0;\n        \n        for (int j = 0; j < s.length(); j++) {\n            // Add the cost of transforming s[j] to t[j] to the current window\n            currWind += abs(s[j] - t[j]);\n            \n            // If the current window cost exceeds maxCost, shrink the window from the left\n            while (currWind > maxCost) {\n                currWind -= abs(s[i] - t[i]);\n                i++;\n            }\n            \n            // Update the maximum length of a valid substring\n            maxLength = max(maxLength, j - i + 1);\n        }\n        \n        return maxLength;\n    }\n};\n           \n```
['String', 'Sliding Window', 'C++', 'Java', 'Python3']