LeetCode 408: Valid Word Abbreviation Solution

Master LeetCode problem 408 (Valid Word Abbreviation), a easy 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.

408. Valid Word Abbreviation

Easy

Two Pointers String

Problem Explanation

Explanation:

Given a non-empty string s and an abbreviation abbr, return whether the string matches the abbreviation.

To solve this problem, we can iterate through both the input string and the abbreviation simultaneously. We compare characters at each position:

If the characters match, we move both pointers forward.
If the character in the abbreviation is a digit, we need to skip that many characters in the input string.
If any mismatch is found or we reach the end of either string or abbreviation prematurely, we return false.

Finally, if we successfully iterate through both strings, we return true.

Time complexity: O(n), where n is the length of the input string.
Space complexity: O(1)

Solution Code

class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        int i = 0, j = 0;
        while (i < word.length() && j < abbr.length()) {
            if (word.charAt(i) == abbr.charAt(j)) {
                i++;
                j++;
            } else if (abbr.charAt(j) >= '1' && abbr.charAt(j) <= '9') {
                int num = 0;
                while (j < abbr.length() && Character.isDigit(abbr.charAt(j))) {
                    num = num * 10 + abbr.charAt(j) - '0';
                    j++;
                }
                i += num;
            } else {
                return false;
            }
        }
        return i == word.length() && j == abbr.length();
    }
}

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