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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

  • 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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