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925. Long Pressed Name

Description

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.

Solutions

• Java
• C++
• Python
• Go

• class Solution {
      public boolean isLongPressedName(String name, String typed) {
          int m = name.length(), n = typed.length();
          int i = 0, j = 0;
          for (; i < m && j < n; ++i, ++j) {
              if (name.charAt(i) != typed.charAt(j)) {
                  return false;
              }
              int cnt1 = 0, cnt2 = 0;
              char c = name.charAt(i);
              while (i + 1 < m && name.charAt(i + 1) == c) {
                  ++i;
                  ++cnt1;
              }
              while (j + 1 < n && typed.charAt(j + 1) == c) {
                  ++j;
                  ++cnt2;
              }
              if (cnt1 > cnt2) {
                  return false;
              }
          }
          return i == m && j == n;
      }
  }
  

• class Solution {
  public:
      bool isLongPressedName(string name, string typed) {
          int m = name.size(), n = typed.size();
          int i = 0, j = 0;
          for (; i < m && j < n; ++i, ++j) {
              if (name[i] != typed[j]) return false;
              int cnt1 = 0, cnt2 = 0;
              char c = name[i];
              while (i + 1 < m && name[i + 1] == c) {
                  ++i;
                  ++cnt1;
              }
              while (j + 1 < n && typed[j + 1] == c) {
                  ++j;
                  ++cnt2;
              }
              if (cnt1 > cnt2) return false;
          }
          return i == m && j == n;
      }
  };
  

• class Solution:
      def isLongPressedName(self, name: str, typed: str) -> bool:
          m, n = len(name), len(typed)
          i = j = 0
          while i < m and j < n:
              if name[i] != typed[j]:
                  return False
              cnt1 = cnt2 = 0
              c = name[i]
              while i + 1 < m and name[i + 1] == c:
                  i += 1
                  cnt1 += 1
              while j + 1 < n and typed[j + 1] == c:
                  j += 1
                  cnt2 += 1
              if cnt1 > cnt2:
                  return False
              i, j = i + 1, j + 1
          return i == m and j == n
  
  

• func isLongPressedName(name string, typed string) bool {
  	m, n := len(name), len(typed)
  	i, j := 0, 0
  	for ; i < m && j < n; i, j = i+1, j+1 {
  		if name[i] != typed[j] {
  			return false
  		}
  		cnt1, cnt2 := 0, 0
  		c := name[i]
  		for i+1 < m && name[i+1] == c {
  			i++
  			cnt1++
  		}
  		for j+1 < n && typed[j+1] == c {
  			j++
  			cnt2++
  		}
  		if cnt1 > cnt2 {
  			return false
  		}
  	}
  	return i == m && j == n
  }
  

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