76. 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.

A substring is a contiguous sequence of characters within the string.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
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 <= 10^5
s and t consist of uppercase and lowercase English letters.

Solution

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
class Solution {
    public String minWindow(String s, String t) {
        int[] freqT = new int[256];
        int[] freqS = new int[256];
        
        for (int i = 0; i < t.length(); i++) {
            freqT[t.charAt(i)]++;
        }
        
        int lo = 0;
        int hi = 0;
        String res = "";
        int min = Integer.MAX_VALUE;
        
        while (hi < s.length()) {
            
            freqS[s.charAt(hi)]++;
            hi++;
            
            while (moreThanEnough(freqT, freqS)) {
                freqS[s.charAt(lo)]--;
                lo++;
                if (hi - lo < min) {
                    res = s.substring(lo - 1, hi);
                    min = hi - lo;
                }
            }
        }
        
        return res;
    }
    
    boolean moreThanEnough(int[] freqT, int[] freqS) {
        for (int i = 0; i < freqS.length; i++) {
            if (freqT[i] == 0) continue;
            if (freqS[i] < freqT[i]) {
                return false;
            }
        }
        return true;
    }
}

Solution#

Solution