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In this post, you will find the solution for the **Minimum Window Substring** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Minimum Window Substring– LeetCode Problem

Minimum Window Substring– LeetCode Problem

**Problem:**

Given 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.length`

`n == t.length`

`1 <= m, n <= 10`

^{5}`s`

and`t`

consist of uppercase and lowercase English letters.

Minimum Window Substring– LeetCode Solutions

Minimum Window Substring in C++:

class Solution { public: string minWindow(string s, string t) { vectorcount(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 in Java:

class 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); } }

Minimum Window Substring in Python:

class 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]