76. Minimum Window Substring

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

For example,
S = "ADOBECODEBANC"
T = "ABC"

Minimum window is "BANC".

Note:
If there is no such window in S that covers all characters in T, return the empty string "".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

Solution:

Method 1:

1. Use two pointers start and end to keep track of the minimum window that satisfy the condition.

2. Move end to find a valid window.

3. If we find a valid window, move start to find a smaller window.



Code:

public class Solution {
    public String minWindow(String s, String t) {
        HashMap<Character, Integer> map = new HashMap<>();
        for (char c : s.toCharArray()) {
            map.put(c, 0);
        }
        for (char c : t.toCharArray()) {
            if (!map.containsKey(c)) {
                map.put(c, 1);
            }
            else {
                map.put(c, map.get(c) + 1);
            }
        }
        
        int start = 0;
        int end = 0;
        int minStart = 0;
        int minLen = Integer.MAX_VALUE;
        int count = t.length();
        while (end < s.length()) {
            char c1 = s.charAt(end);
            if (map.get(c1) > 0) {
                count--;
            }
            map.put(c1, map.get(c1) - 1);
            end++;
            
            while (count == 0) {
                if (end - start < minLen) {
                    minLen = end - start;
                    minStart = start;
                }
                
                char c2 = s.charAt(start);
                map.put(c2, map.get(c2) + 1);
                if (map.get(c2) > 0) {
                    count++;
                }
                
                start++;
            }
        }
        return minLen == Integer.MAX_VALUE ? "" : s.substring(minStart, minStart + minLen);
    }
}

Method 2:


Code:

public class Solution {
    public String minWindow(String s, String t) {
        HashMap<Character, Integer> map = new HashMap<>();
        for (char c : t.toCharArray()) {
            if (!map.containsKey(c)) {
                map.put(c, 1);
            }
            else {
                map.put(c, map.get(c) + 1);
            }
        }
        int len = Integer.MAX_VALUE;
        int head = 0;
        int start = 0; 
        int end = 0;
        int count = map.size();
        while (end < s.length()) {
            char c = s.charAt(end);
            if (map.containsKey(c)) {
                map.put(c, map.get(c) - 1);
                if (map.get(c) == 0) {
                    count--;
                }
            }
            end++;
            while (count == 0) {
                if (end - start < len) {
                    len = end - start;
                    head = start;
                }
                char ch = s.charAt(start);
                if (map.containsKey(ch)) {
                    map.put(ch, map.get(ch) + 1);
                    if (map.get(ch) > 0) {
                        count++;
                    }
                }
                start++;
            }
        }
        if (len == Integer.MAX_VALUE) {
            return "";
        }
        return s.substring(head, head + len);
    }
}