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Solution 16: Sequential Inclusion Span (Minimum Window Subsequence)

Approach Explanation

We use a two-pointer approach with a "backwards check" optimization. When we find a valid subsequence in s1 ending at r, we move backwards from r to find the best (rightmost) starting l for that subsequence. This helps find the minimum window.

Step-by-Step Logic

  1. s1_ptr = 0, s2_ptr = 0, min_len = float('inf'), res = "".
  2. Iterate s1_ptr through s1:
    • If s1[s1_ptr] == s2[s2_ptr], increment s2_ptr.
    • If s2_ptr == len(s2) (found a valid subsequence):
      • Start a back_ptr at s1_ptr and move s2_ptr backwards from len(s2)-1 to 0.
      • Find the starting index l in s1.
      • Update min_len and res.
      • Reset s1_ptr = l + 1, s2_ptr = 0.
  3. Return res.

Complexity

  • Time Complexity: O(N * M) worst case, but efficient in practice.
  • Space Complexity: O(1).

Code

def min_window(s1, s2):
    n1, n2 = len(s1), len(s2)
    s1_idx = 0
    s2_idx = 0
    res = ""
    min_len = float('inf')
    
    while s1_idx < n1:
        if s1[s1_idx] == s2[s2_idx]:
            s2_idx += 1
            if s2_idx == n2:
                # Found a valid end. Go backward to find best start.
                end = s1_idx
                s2_idx -= 1
                while s2_idx >= 0:
                    if s1[s1_idx] == s2[s2_idx]:
                        s2_idx -= 1
                    s1_idx -= 1
                s1_idx += 1
                
                # Check min_len
                if end - s1_idx + 1 < min_len:
                    min_len = end - s1_idx + 1
                    res = s1[s1_idx : end + 1]
                    
                s2_idx = 0
        s1_idx += 1
        
    return res