Problem
Implement regex matching with support for “.” (any single char) and “*” (zero or more of the preceding element).
Example
Input: s = "aab", p = "c*a*b"
Output: trueSolution
2D DP. dp[i][j] = whether s[:i] matches p[:j]. Special handling for * which can match 0 or more chars.
def is_match(s, p):
m, n = len(s), len(p)
dp = [[False] * (n + 1) for _ in range(m + 1)]
dp[0][0] = True
for j in range(1, n + 1):
if p[j-1] == '*': dp[0][j] = dp[0][j-2]
for i in range(1, m + 1):
for j in range(1, n + 1):
if p[j-1] == '*':
dp[i][j] = dp[i][j-2]
if p[j-2] == s[i-1] or p[j-2] == '.':
dp[i][j] = dp[i][j] or dp[i-1][j]
elif p[j-1] == s[i-1] or p[j-1] == '.':
dp[i][j] = dp[i-1][j-1]
return dp[m][n]function isMatch(s, p) {
const m = s.length, n = p.length;
const dp = Array.from({length: m + 1}, () => new Array(n + 1).fill(false));
dp[0][0] = true;
for (let j = 1; j <= n; j++) {
if (p[j-1] === '*') dp[0][j] = dp[0][j-2];
}
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
if (p[j-1] === '*') {
dp[i][j] = dp[i][j-2];
if (p[j-2] === s[i-1] || p[j-2] === '.') dp[i][j] = dp[i][j] || dp[i-1][j];
} else if (p[j-1] === s[i-1] || p[j-1] === '.') {
dp[i][j] = dp[i-1][j-1];
}
}
}
return dp[m][n];
}bool isMatch(string s, string p) {
int m = s.size(), n = p.size();
vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false));
dp[0][0] = true;
for (int j = 1; j <= n; j++) if (p[j-1] == '*') dp[0][j] = dp[0][j-2];
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (p[j-1] == '*') {
dp[i][j] = dp[i][j-2];
if (p[j-2] == s[i-1] || p[j-2] == '.') dp[i][j] = dp[i][j] || dp[i-1][j];
} else if (p[j-1] == s[i-1] || p[j-1] == '.') {
dp[i][j] = dp[i-1][j-1];
}
}
}
return dp[m][n];
}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 j = 1; j <= n; j++) if (p.charAt(j-1) == '*') dp[0][j] = dp[0][j-2];
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
char pc = p.charAt(j-1);
if (pc == '*') {
dp[i][j] = dp[i][j-2];
char prev = p.charAt(j-2);
if (prev == s.charAt(i-1) || prev == '.') dp[i][j] = dp[i][j] || dp[i-1][j];
} else if (pc == s.charAt(i-1) || pc == '.') {
dp[i][j] = dp[i-1][j-1];
}
}
}
return dp[m][n];
}Complexity
- Time: O(m * n)
- Space: O(m * n)
Explanation
For * we have two choices: ignore the pattern pair (zero matches) or use it to match one more char of s. The base case handles patterns like “a*b*” matching empty string.
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