Python Solution with explanation
Intuition\n\nThe problem asks us to find the longest substring where we can change characters in the first string (s) to make it equal to the second string (t) 
## Intuition\n\nThe problem asks us to find the longest substring where we can change characters in the first string (s) to make it equal to the second string (t) within a given cost limit (maxCost). We can solve this using a sliding window approach.\n\n## Approach\n\n1. We\'ll use two pointers: `start` and `end`. `start` will mark the beginning of the current window in string `s`, and `end` will iterate through `s`.\n2. At each step, we\'ll calculate the cost of making the characters at `s[end]` and `t[end]` equal. This cost is the absolute difference between their ASCII values.\n3. We\'ll keep track of the total cost (`current_cost`) for the current window.\n4. If `current_cost` exceeds `maxCost`, we need to reduce the cost. We\'ll achieve this by moving the `start` pointer forward, effectively shrinking the window and removing characters from `s`. We\'ll update `current_cost` accordingly.\n5. As we iterate through `s`, we\'ll keep track of the maximum length (`max_length`) of a valid substring encountered so far.\n\n## Complexity\n\n* **Time complexity:** O(n), where n is the length of the strings. We iterate through each character in `s` at most once.\n* **Space complexity:** O(1). We only use constant extra space for variables like `start`, `end`, `current_cost`, and `max_length`.\n\n# Code\n```\nclass Solution:\n    def equalSubstring(self, s: str, t: str, maxCost: int) -> int:\n        start = 0\n        current_cost = 0\n        max_length = 0\n\n        for end in range(len(s)):\n            # Calculate the cost to change s[end] to t[end]\n            current_cost += abs(ord(s[end]) - ord(t[end]))\n\n            # If the current cost exceeds maxCost, move the start pointer to reduce the cost\n            while current_cost > maxCost:\n                current_cost -= abs(ord(s[start]) - ord(t[start]))\n                start += 1\n\n            # Calculate the maximum length of the valid window\n            max_length = max(max_length, end - start + 1)\n        \n        return max_length\n\n```\n\n## Code Review\n\nThe provided code accurately implements the described approach. Here\'s a breakdown:\n\n1. It initializes `start`, `current_cost`, and `max_length`.\n2. The loop iterates through `s` using `end` as the pointer.\n3. Inside the loop:\n   - It calculates the cost for the current characters `s[end]` and `t[end]`.\n   - It checks if `current_cost` exceeds `maxCost`. If yes, it shrinks the window by moving `start` forward and updates `current_cost`.\n4. Finally, it updates `max_length` if the current window length (`end - start + 1`) is greater.\n5. The function returns `max_length`, which represents the longest valid substring.\n
['Python3']