# LeetCode 76. Minimum Window Substring

## Description

https://leetcode.com/problems/minimum-window-substring/

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?

## Explanation

Use two pointers sliding window. Increase the right pointer to meet the target string, and adjust the left pointer to see where is the place to have a minimum substring to have the target string.

## Python Solution

``````class Solution:
def minWindow(self, s: str, t: str) -> str:

counter_t = {}
counter_s = {}

for c in t:
counter_t[c] = counter_t.get(c, 0) + 1

i = 0
j = 0

left = -1
right = -1

valid = 0

for i in range(len(s)):

while j < len(s) and valid < len(counter_t):
counter_s[s[j]] = counter_s.get(s[j], 0) + 1

if s[j] in counter_t and counter_s[s[j]] == counter_t[s[j]]:
valid += 1

j += 1

if valid == len(counter_t):
if left == -1 or j - i < right - left:
left = i
right = j

counter_s[s[i]] -= 1
if s[i] in counter_t and counter_s[s[i]] == counter_t[s[i]] - 1:
valid -= 1

if left == -1:
return ""

return s[left : right]
``````
• Time Complexity: O(N).
• Space Complexity: O(N).

where Q is the length of queries and N is the length of colors.

## One Thought to “LeetCode 76. Minimum Window Substring”

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