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Formatted question description: https://leetcode.ca/all/2330.html

2330. Valid Palindrome IV

Description

You are given a 0-indexed string s consisting of only lowercase English letters. In one operation, you can change any character of s to any other character.

Return true if you can make s a palindrome after performing exactly one or two operations, or return false otherwise.

 

Example 1:

Input: s = "abcdba"
Output: true
Explanation: One way to make s a palindrome using 1 operation is:
- Change s[2] to 'd'. Now, s = "abddba".
One operation could be performed to make s a palindrome so return true.

Example 2:

Input: s = "aa"
Output: true
Explanation: One way to make s a palindrome using 2 operations is:
- Change s[0] to 'b'. Now, s = "ba".
- Change s[1] to 'b'. Now, s = "bb".
Two operations could be performed to make s a palindrome so return true.

Example 3:

Input: s = "abcdef"
Output: false
Explanation: It is not possible to make s a palindrome using one or two operations so return false.

 

Constraints:

• 1 <= s.length <= 105
• s consists only of lowercase English letters.

Solutions

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public boolean makePalindrome(String s) {
          int cnt = 0;
          int i = 0, j = s.length() - 1;
          for (; i < j; ++i, --j) {
              if (s.charAt(i) != s.charAt(j)) {
                  ++cnt;
              }
          }
          return cnt <= 2;
      }
  }
  

• class Solution {
  public:
      bool makePalindrome(string s) {
          int cnt = 0;
          int i = 0, j = s.size() - 1;
          for (; i < j; ++i, --j) {
              cnt += s[i] != s[j];
          }
          return cnt <= 2;
      }
  };
  

• class Solution:
      def makePalindrome(self, s: str) -> bool:
          i, j = 0, len(s) - 1
          cnt = 0
          while i < j:
              cnt += s[i] != s[j]
              i, j = i + 1, j - 1
          return cnt <= 2
  
  

• func makePalindrome(s string) bool {
  	cnt := 0
  	i, j := 0, len(s)-1
  	for ; i < j; i, j = i+1, j-1 {
  		if s[i] != s[j] {
  			cnt++
  		}
  	}
  	return cnt <= 2
  }
  

• function makePalindrome(s: string): boolean {
      let cnt = 0;
      let i = 0;
      let j = s.length - 1;
      for (; i < j; ++i, --j) {
          if (s[i] != s[j]) {
              ++cnt;
          }
      }
      return cnt <= 2;
  }
  
  

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