LeetCode 1023: Camelcase Matching Solution

Master LeetCode problem 1023 (Camelcase Matching), a medium 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.

1023. Camelcase Matching

Medium

Array Two Pointers String Trie String Matching

Problem Explanation

Explanation

To solve this problem:

Iterate through each query in the queries array.
Initialize two pointers, one for the query string and one for the pattern string.
Check if the characters match between the query and pattern:
- If they match, move both pointers to the next character.
- If they don't match:
  - If the character in the query is uppercase, return false.
  - If the character in the query is lowercase, move only the query pointer.
Check if the pattern pointer reaches the end of the pattern string. If not, return false.
If all characters in the query are matched with the pattern, return true.

Time complexity: O(n * m) where n is the number of queries and m is the maximum length of a query string. Space complexity: O(1) as we are using constant extra space.

Solution Code

class Solution {
    public List<Boolean> camelMatch(String[] queries, String pattern) {
        List<Boolean> result = new ArrayList<>();
        
        for (String query : queries) {
            result.add(isMatch(query, pattern));
        }
        
        return result;
    }
    
    private boolean isMatch(String query, String pattern) {
        int i = 0, j = 0;
        
        while (i < query.length() && j < pattern.length()) {
            if (query.charAt(i) == pattern.charAt(j)) {
                i++;
                j++;
            } else {
                if (Character.isUpperCase(query.charAt(i))) {
                    return false;
                }
                i++;
            }
        }
        
        while (i < query.length()) {
            if (Character.isUpperCase(query.charAt(i))) {
                return false;
            }
            i++;
        }
        
        return j == pattern.length();
    }
}

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