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Wildcard Matching

Certification: Advanced Level Accuracy: 0% Submissions: 0 Points: 15

Write a C program to implement wildcard pattern matching with support for '?' and '*'. The '?' matches any single character and '*' matches any sequence of characters (including empty sequence). The matching should cover the entire input string (not partial).

Example 1
    
  
  • Input: s = "adceb", p = "*a*b*"
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  • Output: true
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  • Explanation: 
    • The pattern "*a*b*" can match "adceb". 
    • First '*' matches "", 'a' matches 'a', second '*' matches "dce", 'b' matches 'b', third '*' matches "". 
    • Therefore, the string matches the pattern.
    •
Example 2
    
  
  • Input: s = "acdcb", p = "a*c?b"
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  • Output: false
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  • Explanation: 
    • The pattern "a*c?b" cannot match "acdcb". 
    • 'a' matches 'a', '*' matches "d", 'c' matches 'c', '?' should match 'd' but we need 'b' at the end. 
    • Since the pattern doesn't fully match, return false.
    •
Constraints
    
  
  • 0 <= s.length, p.length <= 2000
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  • s contains only lowercase English letters
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  • p contains only lowercase English letters, '?' or '*'
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  • Time Complexity: O(m*n) where m and n are lengths of string and pattern
    • 
  
  • Space Complexity: O(m*n)
    •
StringsDynamic Programming CapgeminiSnowflake
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Hint
Solution

Solution Hints

    
  
  • Use dynamic programming to solve this problem efficiently
    • 
  
  • Create a 2D DP table where dp[i][j] represents if first i characters of string match first j characters of pattern
    • 
  
  • Handle base cases: empty string with empty pattern, empty string with pattern containing only '*'
    • 
  
  • For '*', consider two cases: match empty sequence or match one or more characters
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  • For '?', it matches exactly one character
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  • For regular characters, they must match exactly
    •

Steps to solve by this approach:

 Step 1: Create a 2D DP table where dp[i][j] represents if first i characters of string match first j characters of pattern
 Step 2: Initialize base cases - empty string matches empty pattern, and handle patterns starting with '*'
 Step 3: For each cell, if pattern character is '*', it can match empty sequence (dp[i][j-1]) or one/more characters (dp[i-1][j])
 Step 4: If pattern character is '?' or matches string character, check diagonal cell dp[i-1][j-1]
 Step 5: Fill the entire DP table using the above logic
 Step 6: Return dp[m][n] which represents if entire string matches entire pattern
 Step 7: The time complexity is O(m*n) and space complexity is O(m*n)

Submitted Code :