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This approach uses recursion with memoization to solve the problem efficiently. We recursively compare characters of the input string and the pattern, addressing special characters like '.' and '*'. The memoization is used to store results of sub-problems to avoid redundant calculations, which drastically improves the performance.
Time Complexity: O(m * n), where m and n are the lengths of the string and pattern, respectively. We solve every subproblem once and store the result.
Space Complexity: O(m * n) due to recursion stack and memoization storage.
1public class Solution {
2 public boolean isMatch(String s, String p) {
3 Boolean[][] memo = new Boolean[s.length() + 1][p.length() + 1];
4 return dfs(0, 0, s, p, memo);
5 }
6 private boolean dfs(int i, int j, String s, String p, Boolean[][] memo) {
7 if (memo[i][j] != null) {
8 return memo[i][j];
9 }
10 if (j == p.length()) {
11 return i == s.length();
12 }
13 boolean match = i < s.length() && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '.');
14 if (j + 1 < p.length() && p.charAt(j + 1) == '*') {
15 memo[i][j] = dfs(i, j + 2, s, p, memo) || (match && dfs(i + 1, j, s, p, memo));
16 return memo[i][j];
17 }
18 if (match) {
19 memo[i][j] = dfs(i + 1, j + 1, s, p, memo);
20 return memo[i][j];
21 }
22 memo[i][j] = false;
23 return false;
24 }
25}The Java solution employs a DFS strategy enhancing it with memoization via a Boolean matrix to avoid recomputation. For each (i, j) pair of the string and pattern, we decide if they match, handle '*' by skipping it or proceeding in the string.
This approach involves using a DP table to solve the problem by filling up a boolean matrix iteratively. This avoids recursion and stacks overhead, improving performance in certain scenarios. Each cell in the table represents whether the substring of the pattern matches the substring of the input.
Time Complexity: O(m * n) since we traverse the complete matrix.
Space Complexity: O(m * n) due to the DP table used to store subproblem solutions.
1#include <string>
using namespace std;
class Solution {
public:
bool isMatch(string s, string p) {
vector<vector<bool>> dp(s.size() + 1, vector<bool>(p.size() + 1, false));
dp[0][0] = true;
for (int j = 1; j < p.size(); ++j) {
if (p[j] == '*') {
dp[0][j + 1] = dp[0][j - 1];
}
}
for (int i = 0; i < s.size(); ++i) {
for (int j = 0; j < p.size(); ++j) {
if (p[j] == '.' || s[i] == p[j]) {
dp[i + 1][j + 1] = dp[i][j];
} else if (p[j] == '*') {
dp[i + 1][j + 1] = dp[i + 1][j - 1] || (dp[i][j + 1] && (s[i] == p[j - 1] || p[j - 1] == '.'));
}
}
}
return dp[s.size()][p.size()];
}
};This C++ solution leverages tabulation with a matrix employed to store states of string-pattern matches. It iteratively adjusts the table based on character conditions, managing '.' and '*' in the pattern for successful or unmatched cases.