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The dynamic programming approach uses a 2D table to store results of subproblems. We define dp[i][j] as True if the first i characters in the string s can be matched by the first j characters of the pattern p.
Base Cases:
Fill the table using the following rules:
Time Complexity: O(m*n) where m and n are the lengths of s and p, respectively.
Space Complexity: O(m*n) due to the storage of the dp table.
1public class Solution {
2 public boolean isMatch(String s, String p) {
3 int m = s.length();
4 int n = p.length();
5 boolean[][] dp = new boolean[m + 1][n + 1];
6 dp[0][0] = true;
7 for (int j = 1; j <= n; j++) {
8 if (p.charAt(j - 1) == '*') {
9 dp[0][j] = dp[0][j - 1];
10 }
11 }
12 for (int i = 1; i <= m; i++) {
13 for (int j = 1; j <= n; j++) {
14 if (p.charAt(j - 1) == '*') {
15 dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
16 } else {
17 dp[i][j] = dp[i - 1][j - 1] && (p.charAt(j - 1) == '?' || s.charAt(i - 1) == p.charAt(j - 1));
18 }
19 }
20 }
21 return dp[m][n];
22 }
23}
24The core idea is the same dynamic programming approach. The Java implementation makes use of arrays to store intermediate match results for substrings of s and p.
The greedy algorithm approach tracks the position of the last '*' in the pattern and attempts to match remaining characters greedily. We maintain two pointers, one for the string and one for the pattern. When encountering '*', we remember the positions of both pointers in order to backtrack if needed.
If characters in p and s match or if the pattern has '?', both pointers are incremented. For '*', we shift the pattern pointer and remember indices for future potential matches. If there's a mismatch after '*', we use these remembered indices to backtrack.
Time Complexity: O(m + n) due to possible backtracking
Space Complexity: O(1) requiring constant storage for indices.
This greedy algorithm utilizes backtracking facilitated by the last seen '*' position to handle mismatches, allowing us to efficiently manage sequences that can contain many wildcard characters.