LeetCode 10: Regular Expression Matching Solution
Master LeetCode problem 10 (Regular Expression Matching), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
10. Regular Expression Matching
Problem Explanation
Explanation
To solve this problem, we can use dynamic programming. We will create a 2D DP array where dp[i][j] will represent whether the pattern p[0...j-1] matches the string s[0...i-1]. We will then fill the DP array based on the following conditions:
- If p[j-1] is a normal character or '.' (matches any single character), then dp[i][j] is true if dp[i-1][j-1] is true and s[i-1] matches p[j-1].
- If p[j-1] is '*', then we have two cases:
- If p[j-2] matches s[i-1], then dp[i][j] is true if either dp[i][j-2] (zero occurrences of the preceding element) or dp[i-1][j] (one or more occurrences of the preceding element) is true.
Finally, the result will be stored in dp[s.length()][p.length()].
Solution Code
class Solution {
public boolean isMatch(String s, String p) {
int m = s.length(), n = p.length();
boolean[][] dp = new boolean[m + 1][n + 1];
dp[0][0] = true;
for (int i = 0; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (j > 1 && p.charAt(j - 1) == '*') {
dp[i][j] = dp[i][j - 2] || (i > 0 && (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.') && dp[i - 1][j]);
} else {
dp[i][j] = i > 0 && dp[i - 1][j - 1] && (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '.');
}
}
}
return dp[m][n];
}
}Try It Yourself
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