Question

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 $C = 128$.

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]

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);
  }
};

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);
}

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()
  }
}

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);
  }
}