Standard Python moving window (similar problems listed)
Please see and vote for my solutions for these similar problems.\n1208. Get Equal Substrings Within Budget\n3. Longest Substring Without Repeating Characters\n1
Please see and vote for my solutions for these similar problems.\n[1208. Get Equal Substrings Within Budget](https://leetcode.com/problems/get-equal-substrings-within-budget/discuss/392901/Simple-Python-moving-window)\n[3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/348137/Simple-Python-two-pointer-solution-(52ms-beat-97.94))\n[159. Longest Substring with At Most Two Distinct Characters](https://leetcode.com/problems/longest-substring-with-at-most-two-distinct-characters/discuss/348157/Simple-Python-two-pointer-solution)\n[340. Longest Substring with At Most K Distinct Characters](https://leetcode.com/problems/longest-substring-with-at-most-k-distinct-characters/discuss/348216/Simple-Python-two-pointer-solution-(72-ms-beat-94.93))\n[992. Subarrays with K Different Integers](https://leetcode.com/problems/subarrays-with-k-different-integers/discuss/348984/Different-Python-two-pointer-solutions)\n[424. Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement/discuss/363071/Simple-Python-two-pointer-solution)\n[209. Minimum Size Subarray Sum](https://leetcode.com/problems/minimum-size-subarray-sum/discuss/344476/Simple-Python-two-pointer-solution)\n[713. Subarray Product Less Than K](https://leetcode.com/problems/subarray-product-less-than-k/discuss/344245/Simple-Python-solution-(beat-94.59))\n[76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/discuss/344533/Simple-Python-two-pointer-solution)\n\n```\n    def equalSubstring(self, s: str, t: str, maxCost: int) -> int:\n        n = len(s)\n        cost = 0\n        left = 0\n        max_len = 0\n        for right in range(n):\n            cost += abs(ord(s[right]) - ord(t[right]))\n            while cost > maxCost:\n                cost -= abs(ord(s[left]) - ord(t[left]))\n                left += 1\n            max_len = max(max_len, right - left + 1)\n        return max_len\n```
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