In this post, you will know how to solve the Minimum Window Substring Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.Minimum Window Substring Leetcode SolutionsOne more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.ProblemGiven two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "".The testcases will be generated such that the answer is unique.A substring is a contiguous sequence of characters within the string.Example 1:Input: s = "ADOBECODEBANC", t = "ABC" Output: "BANC" Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t. Example 2:Input: s = "a", t = "a" Output: "a" Explanation: The entire string s is the minimum window. Example 3:Input: s = "a", t = "aa" Output: "" Explanation: Both 'a's from t must be included in the window. Since the largest window of s only has one 'a', return empty string. Constraints:m == s.lengthn == t.length1 <= m, n <= 105s and t consist of uppercase and lowercase English letters.Follow up: Could you find an algorithm that runs in O(m + n) time?Minimum Window Substring Leetcode Solutions in Pythonclass Solution: def minWindow(self, s: str, t: str) -> str: count = Counter(t) required = len(t) bestLeft = -1 minLength = len(s) + 1 l = 0 for r, c in enumerate(s): count[c] -= 1 if count[c] >= 0: required -= 1 while required == 0: if r - l + 1 < minLength: bestLeft = l minLength = r - l + 1 count[s[l]] += 1 if count[s[l]] > 0: required += 1 l += 1 return '' if bestLeft == -1 else s[bestLeft: bestLeft + minLength] Minimum Window Substring Leetcode Solutions in CPPclass Solution { public: string minWindow(string s, string t) { vector<int> count(128); int required = t.length(); int bestLeft = -1; int minLength = s.length() + 1; for (const char c : t) ++count[c]; for (int l = 0, r = 0; r < s.length(); ++r) { if (--count[s[r]] >= 0) --required; while (required == 0) { if (r - l + 1 < minLength) { bestLeft = l; minLength = r - l + 1; } if (++count[s[l++]] > 0) ++required; } } return bestLeft == -1 ? "" : s.substr(bestLeft, minLength); } }; Minimum Window Substring Leetcode Solutions in Javaclass Solution { public String minWindow(String s, String t) { int[] count = new int[128]; int required = t.length(); int bestLeft = -1; int minLength = s.length() + 1; for (final char c : t.toCharArray()) ++count[c]; for (int l = 0, r = 0; r < s.length(); ++r) { if (--count[s.charAt(r)] >= 0) --required; while (required == 0) { if (r - l + 1 < minLength) { bestLeft = l; minLength = r - l + 1; } if (++count[s.charAt(l++)] > 0) ++required; } } return bestLeft == -1 ? "" : s.substring(bestLeft, bestLeft + minLength); } } Note: This problem Minimum Window Substring is generated by Leetcode but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.Next: Sort Colors Leetcode Solution Post navigationCombinations Leetcode Solution Sort Colors Leetcode Solution