LeetCode – Minimum Window Subsequence (Java)

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Given strings S and T, find the minimum (contiguous) substring W of S, so that T is a subsequence of W.

If there is no such window in S that covers all characters in T, return the empty string “”. If there are multiple such minimum-length windows, return the one with the left-most starting index.

Example 1:

Input:
S = “abcdebdde”, T = “bde”
Output: “bcde”
Explanation:
“bcde” is the answer because it occurs before “bdde” which has the same length.
“deb” is not a smaller window because the elements of T in the window must occur in order.

Java Solution 1 – Two pointers

In a brute-force way, this problem can be solved by using two pointers which iterate over characters of the two strings respectively.

public String minWindow(String S, String T) {
    int start=0;
    String result = "";
 
    while(start<S.length()){
        int j=0;
 
        for(int i=start; i<S.length(); i++){
            if(S.charAt(i)==T.charAt(j)&&j==0){
                start=i;
            }
 
            if(S.charAt(i)==T.charAt(j)){
                j++;
            }
 
            if(j==T.length()){
                if(result.equals("")||(i-start+1)<result.length()){
                    result = S.substring(start, i+1);
                }
                start=start+1;
                break;
            }
 
            if(i==S.length()-1){
                return result;
            }
        }
    }
 
    return result;
}

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