In this post, we are going 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.

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 <= 105
s
andt
consist of uppercase and lowercase English letters.
Follow up: Could you find an algorithm that runs in O(m + n)
time?
Now, let’s see the leetcode solution of Minimum Window Substring Leetcode Solution.
Minimum Window Substring Leetcode Solution 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]
Minimum Window Substring Leetcode Solution in CPP
class 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 Solution 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); } }
Note: This problem Minimum Window Substring is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.