727. Minimum Window Subsequence

Given strings s1 and s2, return the minimum contiguous substring part of s1, so that s2 is a subsequence of the part.

If there is no such window in s1 that covers all characters in s2, return the empty string "". If there are multiple such minimum-length windows, return the one with the left-most starting index.

Example 1:

Input: s1 = "abcdebdde", s2 = "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 s2 in the window must occur in order.

Example 2:

Input: s1 = "jmeqksfrsdcmsiwvaovztaqenprpvnbstl", s2 = "u"
Output: ""

class Solution {
    public String minWindow(String S, String T) {
        if (S.isEmpty() || T.isEmpty()) {
            return "";
        }

        int right = 0, minLen = Integer.MAX_VALUE;
        String result = "";

        while (right < S.length()) {
            int tIndex = 0;
            while (right < S.length()) {
                if (S.charAt(right) == T.charAt(tIndex)) {
                    tIndex++;
                }
                if (tIndex == T.length()) {
                    break;
                }
                right++;
            }

            // if right pointer is over than boundary
            if (right == S.length()) {
                break;
            }

            // use another slow pointer to traverse from right to left until find first
            // character of T in S
            int left = right;
            tIndex = T.length() - 1;
            while (left >= 0) {
                if (S.charAt(left) == T.charAt(tIndex)) {
                    tIndex--;
                }
                if (tIndex == -1) {
                    break;
                }
                left--;
            }
            // if we found another subsequence with smaller length, update result
            if (right - left + 1 < minLen) {
                minLen = right - left + 1;
                result = S.substring(left, right + 1);
            }
            right = left + 1;
        }
        return result;
    }
}