# Minimum Window Substring Leetcode Solution

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.

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` and `t` consist of uppercase and lowercase English letters.

Follow up: Could you find an algorithm that runs in `O(m + n)` time?

Now, lets 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 SubstringLeetcode Solutionin 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 SubstringLeetcode 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.

Sharing Is Caring