MANG Problem #23: Minimum Window Subsequence (Hard)

1. Problem Statement

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

If there is no such window in S that contains all characters in T after keeping the order of characters in T, return an empty string "". If there are multiple such windows of the same minimum length, return the one with the smallest starting index.

Input: S = "abcdebdde", T = "bde"
Output: "bcde" (Not "bdde", as "bcde" appears earlier)

2. Approach: Sliding Window + Reverse Verification

Standard sliding window doesn't work directly because we need to maintain order. We use a two-pass optimization for every potential window.

Find a Valid Window:
- Iterate through S with a pointer i.
- Maintain a pointer j for T.
- If S[i] == T[j], increment j.
- When j reaches the end of T, we found a potential window ending at i.
Optimize the Start:
- We know the window ends at i. But where should it start to be as small as possible?
- Move backward from i to find the last occurrence of each character of T in reverse order.
- This gives us the optimal (latest) start index for the current end index.
Update: Record the smallest window and reset j to 0. Restart the search from start + 1.

3. Java Implementation

public String minWindow(String S, String T) {
    char[] s = S.toCharArray(), t = T.toCharArray();
    int i = 0, j = 0;
    int minLen = Integer.MAX_VALUE;
    String res = "";

    while (i < s.length) {
        if (s[i] == t[j]) {
            j++;
            // Found a valid window
            if (j == t.length) {
                int end = i;
                j--;
                // Pass 2: Shrink from right to left to find optimal start
                while (j >= 0) {
                    if (s[i] == t[j]) j--;
                    i--;
                }
                i++; // Pushed back to the start index
                j++; // Reset j to 0 for next search
                
                if (end - i + 1 < minLen) {
                    minLen = end - i + 1;
                    res = S.substring(i, end + 1);
                }
                // Important: Jump back to start + 1 to find the next potential window
                i = i + 1;
                j = 0;
            }
        }
        i++;
    }
    return res;
}

4. 5-Minute "Video-Style" Walkthrough

The "Aha!" Moment: If you find "b...d...e", you've found a window. But the "b" you found at the start might not be the best "b". There might be another "b" much closer to the "d".
The Reverse Pass: By walking backward from the "e" we just found, we guarantee we find the latest possible "d" and "b" that still form the sequence. This "squeeze" from both sides ensures the window is as small as possible.
The Index Jump: After finding a window, don't just continue from i. Jump back to start + 1. This handles overlapping windows like S="bbde", T="bde".

5. Interview Discussion

Interviewer: "What is the time complexity?"
You: "In the worst case, $O(S^2)$, but in practice, it is closer to $O(S)$ because we only perform the reverse pass when a full match is found."
Interviewer: "Can we use DP?"
You: "Yes! We can use a 2D array where dp[i][j] is the starting index of the shortest substring of S[0...j] that contains T[0...i] as a subsequence. This would be a strict $O(S \times T)$ solution."