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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

• 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;
}