Valid Palindrome IV

Valid Palindrome IV - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: s = "abcdba"

› Output: true

💡 Note: Compare from ends: a=a (match), b=b (match), c≠d (1 mismatch). Total mismatches = 1 ≤ 2, so return true.

Example 2 — Two Mismatches

$ Input: s = "aa"

› Output: true

💡 Note: Already a palindrome (0 mismatches), but we need exactly 1-2 operations. We can change any character to make 1 operation.

Example 3 — Too Many Mismatches

$ Input: s = "abcdef"

› Output: false

💡 Note: Compare pairs: a≠f, b≠e, c≠d. That's 3 mismatches > 2, so impossible with only 1-2 operations.

Constraints

1 ≤ s.length ≤ 10⁵
s consists of only lowercase English letters

Visualization

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Asked in

F Facebook 35 a Amazon 28 M Microsoft 22

The key insight is that to make a palindrome with 1-2 operations, we need at most 2 character mismatches when comparing from both ends. Best approach uses two pointers to count mismatches in O(n) time. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Try All Changes

⏱️ Time: O(n³) Space: O(n)

Generate all possible strings by changing 1 or 2 characters, then check if any resulting string is a palindrome. This involves trying every position and every possible character replacement.

Two Pointers - Count Mismatches

⏱️ Time: O(n) Space: O(1)

Place pointers at start and end, move inward comparing characters. Count how many pairs don't match - if ≤2 mismatches, we can fix with ≤2 operations to make palindrome.

Brute Force - Try All Changes — Algorithm Steps

Step 1: Try changing each single character to all 26 possible letters
Step 2: Try changing each pair of characters to all possible combinations
Step 3: For each modified string, check if it's a palindrome
Step 4: Return true if any palindrome is found

Visualization

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Step-by-Step Walkthrough

Try 1 Change

Test each position with all 26 letters

Try 2 Changes

Test each pair of positions with all letter combinations

Check Palindrome

For each modification, verify if result is palindrome

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

bool isPalindrome(const char* str) {
    int n = strlen(str);
    for (int i = 0; i < n/2; i++) {
        if (str[i] != str[n-1-i]) {
            return false;
        }
    }
    return true;
}

bool solution(char* s) {
    int n = strlen(s);
    char modified[1001];
    
    // Try changing exactly 1 character
    for (int i = 0; i < n; i++) {
        for (char c = 'a'; c <= 'z'; c++) {
            if (c != s[i]) {
                strcpy(modified, s);
                modified[i] = c;
                if (isPalindrome(modified)) {
                    return true;
                }
            }
        }
    }
    
    // Try changing exactly 2 characters
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            for (char c1 = 'a'; c1 <= 'z'; c1++) {
                for (char c2 = 'a'; c2 <= 'z'; c2++) {
                    if (c1 != s[i] || c2 != s[j]) {
                        strcpy(modified, s);
                        modified[i] = c1;
                        modified[j] = c2;
                        if (isPalindrome(modified)) {
                            return true;
                        }
                    }
                }
            }
        }
    }
    
    return false;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    bool result = solution(s);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

O(n²) possible changes × O(n) palindrome check for each

⚠ Quadratic Growth

Space Complexity

O(n)

Storing modified string copies during checking

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try All Changes — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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