LeetCode-in-All

76. Minimum Window Substring

Hard

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.

A 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?

Solution

impl Solution {
    pub fn min_window(s: String, t: String) -> String {
        let mut map = vec![0; 128];
        for ch in t.chars() {
            map[ch as usize] += 1;
        }

        let mut count = t.len();
        let (mut begin, mut end, mut head) = (0, 0, 0);
        let mut d = usize::MAX;

        let s_chars: Vec<char> = s.chars().collect();
        while end < s.len() {
            if map[s_chars[end] as usize] > 0 {
                count -= 1;
            }
            map[s_chars[end] as usize] -= 1;
            end += 1;

            while count == 0 {
                if end - begin < d {
                    d = end - begin;
                    head = begin;
                }
                if map[s_chars[begin] as usize] == 0 {
                    count += 1;
                }
                map[s_chars[begin] as usize] += 1;
                begin += 1;
            }
        }

        if d == usize::MAX {
            return "".to_string();
        }
        s_chars[head..head + d].iter().collect()
    }
}

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