Your friend is typing his name into a keyboard. Sometimes, when typing a character c, the key might get long pressed, and the character will be typed 1 or more times.

You examine the typed characters of the keyboard. Return True if it is possible that it was your friends name, with some characters (possibly none) being long pressed.

Example 1:

Input: name = "alex", typed = "aaleex" Output: true Explanation: 'a' and 'e' in 'alex' were long pressed. Example 2:

Input: name = "saeed", typed = "ssaaedd" Output: false Explanation: 'e' must have been pressed twice, but it was not in the typed output.

Constraints:

1 <= name.length, typed.length <= 1000 name and typed consist of only lowercase English letters.

SOLUTION:

        if(name.length() > typed.length()){
            return false;
        }

        int i = 0;
        int j = 0;

        int nLen = name.length();
        int tLen = typed.length();

        while(i < nLen && j < tLen){
            char ch = name.charAt(i);

            if(name.charAt(i) != typed.charAt(j)){
                return false;
            }
            i++; j++;

            int nameCharLen = 0;
            int typedCharLen = 0;

            while(i < nLen && ch == name.charAt(i)){
                i++;
                nameCharLen++;
            }

            while(j < tLen && ch == typed.charAt(j)){
                j++;
                typedCharLen++;
            }

            if(nameCharLen > typedCharLen){
                return false;
            }
        }

        if(i == nLen && j == tLen){
            return true;
        }

        return false;
    }
}

Runtime: 1 ms, faster than 92.23% of Java online submissions for Long Pressed Name. Memory Usage: 42.2 MB, less than 42.46% of Java online submissions for Long Pressed Name.