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Hard
Hash Table
String
Sliding Window

76. Minimum Window Substring

中文文档

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 $C = 128$.

Python3

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        need = Counter(t)
        window = Counter()
        cnt, j, k, mi = 0, 0, -1, inf
        for i, c in enumerate(s):
            window[c] += 1
            if need[c] >= window[c]:
                cnt += 1
            while cnt == len(t):
                if i - j + 1 < mi:
                    mi = i - j + 1
                    k = j
                if need[s[j]] >= window[s[j]]:
                    cnt -= 1
                window[s[j]] -= 1
                j += 1
        return '' if k < 0 else s[k : k + mi]

Java

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);
    }
}

C++

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);
    }
};

Go

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]
}

TypeScript

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);
}

Rust

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()
    }
}

C#

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);
    }
}