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Solution 16: Permissive Pattern Matcher (Wildcard Matching)

Approach Explanation

This can be solved with DP or a greedy two-pointer approach with backtracking. The greedy approach is faster for wildcard matching.

Step-by-Step Logic

  1. s_ptr = 0, p_ptr = 0, star_ptr = -1, s_tmp = -1.
  2. While s_ptr < len(s):
    • If p_ptr < len(p) and p[p_ptr] is s[s_ptr] or ?: Increment both.
    • Else if p_ptr < len(p) and p[p_ptr] is *: Mark star = p_ptr, tmp_s = s_ptr. Increment p_ptr.
    • Else if star_ptr != -1: (Mismatch but we saw a *): Backtrack p_ptr = star_ptr + 1, s_tmp += 1, s_ptr = s_tmp.
    • Else: return False.
  3. Check remaining p characters: must all be *.

Complexity

  • Time Complexity: O(N * M) worst case, but near O(N + M) average.
  • Space Complexity: O(1).

Code

def is_wildcard_match(s, p):
    s_ptr = p_ptr = 0
    star_idx = s_tmp_idx = -1
    
    while s_ptr < len(s):
        # Case 1: Match or ?
        if p_ptr < len(p) and (p[p_ptr] == '?' or p[p_ptr] == s[s_ptr]):
            s_ptr += 1
            p_ptr += 1
        # Case 2: *
        elif p_ptr < len(p) and p[p_ptr] == '*':
            star_idx = p_ptr
            s_tmp_idx = s_ptr
            p_ptr += 1
        # Case 3: Mismatch, use last *
        elif star_idx != -1:
            p_ptr = star_idx + 1
            s_tmp_idx += 1
            s_ptr = s_tmp_idx
        else:
            return False
            
    # Trailing * in p
    return all(x == '*' for x in p[p_ptr:])