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Java-based LeetCode algorithm problem solutions, regularly updated
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package g0001_0100.s0044_wildcard_matching;

// #Hard #Top_Interview_Questions #String #Dynamic_Programming #Greedy #Recursion
// #Udemy_Dynamic_Programming #2023_08_11_Time_2_ms_(99.87%)_Space_43.2_MB_(99.49%)

/**
 * 44 - Wildcard Matching\.
 *
 * Hard
 *
 * Given an input string (`s`) and a pattern (`p`), implement wildcard pattern matching with support for `'?'` and `'*'` where:
 *
 * *   `'?'` Matches any single character.
 * *   `'*'` Matches any sequence of characters (including the empty sequence).
 *
 * The matching should cover the **entire** input string (not partial).
 *
 * **Example 1:**
 *
 * **Input:** s = "aa", p = "a"
 *
 * **Output:** false
 *
 * **Explanation:** "a" does not match the entire string "aa". 
 *
 * **Example 2:**
 *
 * **Input:** s = "aa", p = "\*"
 *
 * **Output:** true
 *
 * **Explanation:** '\*' matches any sequence. 
 *
 * **Example 3:**
 *
 * **Input:** s = "cb", p = "?a"
 *
 * **Output:** false
 *
 * **Explanation:** '?' matches 'c', but the second letter is 'a', which does not match 'b'. 
 *
 * **Example 4:**
 *
 * **Input:** s = "adceb", p = "\*a\*b"
 *
 * **Output:** true
 *
 * **Explanation:** The first '\*' matches the empty sequence, while the second '\*' matches the substring "dce". 
 *
 * **Example 5:**
 *
 * **Input:** s = "acdcb", p = "a\*c?b"
 *
 * **Output:** false 
 *
 * **Constraints:**
 *
 * *   `0 <= s.length, p.length <= 2000`
 * *   `s` contains only lowercase English letters.
 * *   `p` contains only lowercase English letters, `'?'` or `'*'`.
**/
public class Solution {
    public boolean isMatch(String inputString, String pattern) {
        int i = 0;
        int j = 0;
        int starIdx = -1;
        int lastMatch = -1;
        while (i < inputString.length()) {
            if (j < pattern.length()
                    && (inputString.charAt(i) == pattern.charAt(j) || pattern.charAt(j) == '?')) {
                i++;
                j++;
            } else if (j < pattern.length() && pattern.charAt(j) == '*') {
                starIdx = j;
                lastMatch = i;
                j++;
            } else if (starIdx != -1) {
                // there is a no match and there was a previous star, we will reset the j to indx
                // after star_index
                // lastMatch will tell from which index we start comparing the string if we
                // encounter * in pattern
                j = starIdx + 1;
                // we are saying we included more characters in * so we incremented the
                lastMatch++;
                // index
                i = lastMatch;
            } else {
                return false;
            }
        }
        boolean isMatch = true;
        while (j < pattern.length() && pattern.charAt(j) == '*') {
            j++;
        }
        if (i != inputString.length() || j != pattern.length()) {
            isMatch = false;
        }
        return isMatch;
    }
}