Long Pressed Name

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Your friend typed your name but some characters may have been pressed too long. Given the name you intended and the typed string, return true if typed could be a long-pressed version of name.

Example 1
Input: name="alex", typed="aaleex" → Output: true
Example 2
Input: name="saeed", typed="ssaaedd" → Output: false

Solution Approach

Two pointers: match each character of name with typed. Consecutive duplicates in typed are allowed.

  • i pointer on name, j pointer on typed.
  • While j < typed.length:
  • If name[i]==typed[j]: advance both.
  • Else if j>0 and typed[j]==typed[j-1]: advance j (long press).
  • Else return false.
  • Return i == name.length.
Java
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();
    }
}
  • Time Complexity: O(N+M)
  • Space Complexity: O(1)