2330. Valid Palindrome IV

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.

Solution:

class Solution {
    public boolean makePalindrome(String s) {
        int left = 0;
        int right = s.length() -1;

        int count = 0;


        while(left <= right){
            if (s.charAt(left) == s.charAt(right)){
                left++;
                right--;
            }else {
                if (count == 1){
                    return isParlindrome(s, left + 1, right -1);
                }else if (count == 0){
                    if (isParlindrome(s, left + 1, right - 1)){
                        return true;
                    }else{
                        count = 1;
                        left++;
                        right--;
                        continue;
                    }
                }
            }
        }

        return true;
    }

    private boolean isParlindrome(String s, int left, int right){
        while(left <= right){
            if (s.charAt(left) == s.charAt(right)){
                left++;
                right--;
            }else{
                return false;
            }
        }

        return true;
    }
}

// TC: O(n)
// SC: O(n)