# 76. Minimum Window Substring

## Description

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.

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?

## Solutions

Solution 1: Counting + Two Pointers

We use a hash table or array $need$ to count the number of occurrences of each character in string $t$, and another hash table or array $window$ to count the number of occurrences of each character in the sliding window. In addition, we define two pointers $j$ and $i$ to point to the left and right boundaries of the window, respectively. The variable $cnt$ represents how many characters in $t$ are already included in the window. The variables $k$ and $mi$ represent the starting position and length of the minimum covering substring, respectively.

We traverse the string $s$ from left to right. For the currently traversed character $s[i]$:

We add it to the window, i.e., $window[s[i]] = window[s[i]] + 1$. If $need[s[i]] \geq window[s[i]]$ at this time, it means that $s[i]$ is a “necessary character”, so we increment $cnt$ by one. If $cnt$ equals the length of $t$, it means that all characters in $t$ are already included in the window at this time, so we can try to update the starting position and length of the minimum covering substring. If $i - j + 1 \lt mi$, it means that the substring represented by the current window is shorter, so we update $mi = i - j + 1$ and $k = j$. Then, we try to move the left boundary $j$. If $need[s[j]] \geq window[s[j]]$ at this time, it means that $s[j]$ is a “necessary character”. When moving the left boundary, the character $s[j]$ will be removed from the window, so we need to decrement $cnt$ by one, then update $window[s[j]] = window[s[j]] - 1$, and move $j$ one step to the right. If $cnt$ does not equal the length of $t$, it means that all characters in $t$ are not yet included in the window at this time, so we don’t need to move the left boundary, just move $i$ one step to the right and continue to traverse.

After the traversal, if the minimum covering substring is not found, return an empty string, otherwise return $s[k:k+mi]$.

The time complexity is $O(m + n)$, and the space complexity is $O(C)$. Here, $m$ and $n$ are the lengths of strings $s$ and $t$ respectively; and $C$ is the size of the character set, in this problem $C = 128$.

### why <= is used instead of < in the condition if d[c] <= need[c]: cnt += 1?

Let’s consider an example to illustrate why <= is used instead of < in the condition if d[c] <= need[c]: cnt += 1.

Suppose we have:

• s = "ABAACBAB"
• t = "ABC"

We want to find the minimum window in s that contains all the characters in t.

### Step-by-Step Process:

1. Initialization:
• need = {'A': 1, 'B': 1, 'C': 1} (The count of each character needed from t.)
• d = {} (To keep track of characters in the current window.)
• cnt = 0 (To count how many of the required characters we have in the current window.)
2. First Window:
• We traverse s and encounter A. d = {'A': 1}.
• Since d['A'] (1) <= need['A'] (1), we increment cnt to 1.
• We continue and encounter another A. Now, d = {'A': 2}.
• Now, d['A'] (2) > need['A'] (1), so we don’t increment cnt. We have more As than needed for a valid window.
3. Continuing:
• The next character is B. d = {'A': 2, 'B': 1}.
• d['B'] (1) <= need['B'] (1), so we increment cnt to 2.
• Then, we encounter A again, d = {'A': 3, 'B': 1}, still no increment to cnt because we already have more As than required.
• Next is C. d = {'A': 3, 'B': 1, 'C': 1}.
• d['C'] (1) <= need['C'] (1), increment cnt to 3.
4. Valid Window Found:
• At this point, cnt == len(t), meaning we have a valid window that contains at least as many of each required character as in t.
• If we used < instead of <=, we wouldn’t have counted the first occurrence of each character towards cnt when it exactly matched the requirement, potentially missing the earliest valid window.
5. Shrinking the Window:
• As we continue to traverse and adjust our window, we aim to minimize it while maintaining cnt == len(t).
• The condition d[c] <= need[c] correctly increments cnt for characters that meet the requirement exactly, crucial for identifying the minimum window.

In this example, using <= allows us to correctly increment cnt when a character occurrence transitions from not meeting to meeting the exact requirement. It ensures we recognize when we first have a valid window ("ABAAC"), allowing us to then attempt to minimize it while still covering all characters in t. The use of < would delay recognizing a valid window until we had more characters than necessary, potentially missing the minimum valid window.

• class Solution {
public String minWindow(String s, String t) {
int[] need = new int[128];
int[] window = new int[128];
int m = s.length(), n = t.length();
for (int i = 0; i < n; ++i) {
++need[t.charAt(i)];
}
int cnt = 0, j = 0, k = -1, mi = 1 << 30;
for (int i = 0; i < m; ++i) {
++window[s.charAt(i)];
if (need[s.charAt(i)] >= window[s.charAt(i)]) {
++cnt;
}
while (cnt == n) {
if (i - j + 1 < mi) {
mi = i - j + 1;
k = j;
}
if (need[s.charAt(j)] >= window[s.charAt(j)]) {
--cnt;
}
--window[s.charAt(j++)];
}
}
return k < 0 ? "" : s.substring(k, k + mi);
}
}

• class Solution {
public:
string minWindow(string s, string t) {
int need[128]{};
int window[128]{};
int m = s.size(), n = t.size();
for (char& c : t) {
++need[c];
}
int cnt = 0, j = 0, k = -1, mi = 1 << 30;
for (int i = 0; i < m; ++i) {
++window[s[i]];
if (need[s[i]] >= window[s[i]]) {
++cnt;
}
while (cnt == n) {
if (i - j + 1 < mi) {
mi = i - j + 1;
k = j;
}
if (need[s[j]] >= window[s[j]]) {
--cnt;
}
--window[s[j++]];
}
}
return k < 0 ? "" : s.substr(k, mi);
}
};

• '''
>>> deq = collections.deque([])
>>> deq.append(11)
>>> deq.append(22)
>>> deq.append(33)
>>>
>>> deq[0]
11
'''

import collections

class Solution:
def minWindow(self, s: str, t: str) -> str:
cnt = 0
need = collections.Counter(t)
start, end = len(s), 3 * len(s) # Arbitrary 3, just make sure end-start is larger than input s length, for later min-check
d = {}
# Using queue to store indexes, good for a large amount of API calls
deq = collections.deque([])
for i, c in enumerate(s):
if c in need:
deq.append(i)
d[c] = d.get(c, 0) + 1
# '=' also +1, because it's inceased already one line above :)
if d[c] <= need[c]:
cnt += 1
while deq and d[s[deq[0]]] > need[s[deq[0]]]:
d[s[deq.popleft()]] -= 1
if cnt == len(t) and deq[-1] - deq[0] < end - start:
start, end = deq[0], deq[-1]
return s[start:end + 1]

############

from collections import Counter

class Solution:
def minWindow(self, s: str, t: str) -> str:
ans = ''
m, n = len(s), len(t)
if m < n:
return ans
need = Counter(t)
window = Counter()
i, cnt, mi = 0, 0, inf
for j, c in enumerate(s):
window[c] += 1
if need[c] >= window[c]: # >= , because 1 line above already +=1
cnt += 1
while cnt == n:
if j - i + 1 < mi: # in while
mi = j - i + 1
ans = s[i : j + 1]
c = s[i]
if need[c] >= window[c]: # char in window but not in need, need[c]=0, window[c]=1..2..
cnt -= 1
window[c] -= 1
i += 1
return ans


• func minWindow(s string, t string) string {
need := [128]int{}
window := [128]int{}
for _, c := range t {
need[c]++
}
cnt, j, k, mi := 0, 0, -1, 1<<30
for i, c := range s {
window[c]++
if need[c] >= window[c] {
cnt++
}
for cnt == len(t) {
if i-j+1 < mi {
mi = i - j + 1
k = j
}
if need[s[j]] >= window[s[j]] {
cnt--
}
window[s[j]]--
j++
}
}
if k < 0 {
return ""
}
return s[k : k+mi]
}

• function minWindow(s: string, t: string): string {
const need: number[] = new Array(128).fill(0);
const window: number[] = new Array(128).fill(0);
for (const c of t) {
++need[c.charCodeAt(0)];
}
let cnt = 0;
let j = 0;
let k = -1;
let mi = 1 << 30;
for (let i = 0; i < s.length; ++i) {
++window[s.charCodeAt(i)];
if (need[s.charCodeAt(i)] >= window[s.charCodeAt(i)]) {
++cnt;
}
while (cnt === t.length) {
if (i - j + 1 < mi) {
mi = i - j + 1;
k = j;
}
if (need[s.charCodeAt(j)] >= window[s.charCodeAt(j)]) {
--cnt;
}
--window[s.charCodeAt(j++)];
}
}
return k < 0 ? '' : s.slice(k, k + mi);
}


• public class Solution {
public string MinWindow(string s, string t) {
int[] need = new int[128];
int[] window = new int[128];
foreach (var c in t) {
++need[c];
}
int cnt = 0, j = 0, k = -1, mi = 1 << 30;
for (int i = 0; i < s.Length; ++i) {
++window[s[i]];
if (need[s[i]] >= window[s[i]]) {
++cnt;
}
while (cnt == t.Length) {
if (i - j + 1 < mi) {
mi = i - j + 1;
k = j;
}
if (need[s[j]] >= window[s[j]]) {
--cnt;
}
--window[s[j++]];
}
}
return k < 0 ? "" : s.Substring(k, mi);
}
}

• impl Solution {
pub fn min_window(s: String, t: String) -> String {
let (mut need, mut window, mut cnt) = ([0; 256], [0; 256], 0);
for c in t.chars() {
need[c as usize] += 1;
}
let (mut j, mut k, mut mi) = (0, -1, 1 << 31);
for (i, c) in s.chars().enumerate() {
window[c as usize] += 1;
if need[c as usize] >= window[c as usize] {
cnt += 1;
}

while cnt == t.len() {
if i - j + 1 < mi {
k = j as i32;
mi = i - j + 1;
}
let l = s.chars().nth(j).unwrap() as usize;
if need[l] >= window[l] {
cnt -= 1;
}
window[l] -= 1;
j += 1;
}
}
if k < 0 {
return "".to_string();
}
let k = k as usize;
s[k..k + mi].to_string()
}
}