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2330. Valid Palindrome IV πŸ”’

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

Solution 1

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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

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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;
    }
}

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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;
    }
};

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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
}

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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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