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Help this newbie solve an easy problem
By StellarSpecter, history, 3 months ago, In English

Problem: Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:

'.' Matches any single character.​​​​

'*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

Can anyone help me with this ?

i tried two pointers and it failed, then I looked at the solution. It was DP but I could not get the idea on how to think and understand that, Can anyone explain in easiest words to me please ?

Thanks

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3 months ago, # |
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I'd be happy to help you understand the dynamic programming approach for regular expression matching.

Key idea:

The problem can be broken down into two main cases:

  1. The current character in the pattern (p) matches the current character in the input string (s).
  2. The current character in the pattern is '*'.

The goal is to determine whether the entire pattern matches the entire input string.

Dynamic Programming (DP) approach:

Create a 2D array dp with dimensions (len(s) + 1) x (len(p) + 1) where dp[i][j] represents whether the first i characters in s match the first j characters in p.

Initialization:

  • dp[0][0] = true (empty pattern matches empty string)
  • dp[0][j] = false for j > 0 (pattern is not empty, but string is empty)
  • dp[i][0] = false for i > 0 (string is not empty, but pattern is empty)

Transition rules:

  1. If p[j - 1] == '.', then dp[i][j] = dp[i - 1][j - 1] (dot matches any single character)
  2. If p[j - 1] == s[i - 1], then dp[i][j] = dp[i - 1][j - 1] (match if characters match)
  3. If p[j - 1] == '*', then:
    • If dp[i][j - 2] is true, then dp[i][j] = true (zero or more of the preceding element)
    • If p[j - 2] == s[i - 1] || p[j - 2] == '.', then dp[i][j] = dp[i][j - 2] || dp[i - 1][j] (match if zero or more preceding elements match)

Return value:

dp[len(s)][len(p)] represents whether the entire pattern matches the entire input string.

Example:

Let's consider the input string "aab" and pattern "c*a*b".

i dp[i][j]
0 [false, false, false, false, ...]
1 [false, false, false, true, ...]
2 [true, false, false, true, ...]
...

The final result is dp[3][4] = true, indicating that the pattern "c*a*b" matches the input string "aab".

Why it works:

The DP approach ensures that we consider all possible combinations of matching and non-matching characters in both the input string and the pattern. By using the transition rules and initialization values, we can efficiently determine whether the entire pattern matches the entire input string.

I hope this explanation helps you understand the dynamic programming approach for regular expression matching!

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3 months ago, # |
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Auto comment: topic has been updated by StellarSpecter (previous revision, new revision, compare).

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3 months ago, # |
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Let me explain:

IF CHARACTER IS "." then. string must have single character at some index

else IF CHARACTER IS "*" therefore: it will equal previous or not element.

THEREFORE "." character will always work because it matches any character.

SIMILARLY "*" character wiill always work because it need not be equal previous character

Therefore need for dynamic programming is unnecessary. I think entire string works.

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