LeetCode 925: Long Pressed Name Solution

Master LeetCode problem 925 (Long Pressed Name), 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.

925. Long Pressed Name

Easy

Two Pointers String

Problem Explanation

Explanation

To solve this problem, we can iterate through both the name and typed strings simultaneously. We keep track of the current characters being compared in both strings. If the characters match, we move to the next characters in both strings. If they don't match, we check if the current character in typed is the same as the previous character in typed. If it is, this means the current character in typed might have been long pressed, so we move to the next character in typed and continue comparing. At the end, if we reach the end of both strings, it means the typed string can be formed by long-pressing the name string.

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

Solution Code

class Solution {
    public boolean isLongPressedName(String name, String typed) {
        int i = 0, j = 0;
        while (j < typed.length()) {
            if (i < name.length() && name.charAt(i) == typed.charAt(j)) {
                i++;
                j++;
            } else if (j > 0 && typed.charAt(j) == typed.charAt(j - 1)) {
                j++;
            } else {
                return false;
            }
        }
        return i == name.length();
    }
}

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