Valid Palindrome II

Valid Palindrome II - Problem

Given a string s, return true if the string can be a palindrome after deleting at most one character from it.

A palindrome reads the same forward and backward. For example, "racecar" and "level" are palindromes.

Your task is to determine if removing zero or one character can make the string a palindrome.

Input & Output

Example 1 — Can Make Palindrome

$ Input: s = "aba"

› Output: true

💡 Note: "aba" is already a palindrome, so we return true without deleting any character

Example 2 — Need One Deletion

$ Input: s = "abca"

› Output: true

💡 Note: We can delete 'c' to get "aba" or delete 'b' to get "aca", both are palindromes

Example 3 — Cannot Make Palindrome

$ Input: s = "abc"

› Output: false

💡 Note: No matter which character we delete, we cannot make it a palindrome with just one deletion

Constraints

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

Visualization

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Asked in

F Facebook 45 M Microsoft 38 a Amazon 32 G Google 28

The key insight is to use two pointers from both ends. When characters match, move inward. When they mismatch, try skipping either the left or right character and check if the remaining substring is a palindrome. Best approach is two pointers with Time: O(n), Space: O(1).

Common Approaches

✓ Brute Force - Try All Deletions

⏱️ Time: O(n²) Space: O(n)

For each position in the string, create a new string without that character and check if it's a palindrome. Also check if the original string is already a palindrome.

Two Pointers - Optimized Approach

⏱️ Time: O(n) Space: O(1)

Start with two pointers from both ends. When characters match, move both pointers inward. When they don't match, try skipping either the left or right character and check if remaining substring is palindrome.

Brute Force - Try All Deletions — Algorithm Steps

Check if original string is palindrome
For each position i, remove character at i
Check if resulting string is palindrome
Return true if any case works

Visualization

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Step-by-Step Walkthrough

Original String

Check if 'raceacar' is palindrome

Try Deletions

Remove each character and test

Find Solution

Remove middle 'a' gives 'racecar' - palindrome!

Code -

solution.c — C

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

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

bool solution(char* s) {
    int len = strlen(s);
    
    // Check if already palindrome
    if (isPalindrome(s)) {
        return true;
    }
    
    // Try removing each character
    char* temp = (char*)malloc((len) * sizeof(char));
    for (int i = 0; i < len; i++) {
        int k = 0;
        for (int j = 0; j < len; j++) {
            if (j != i) {
                temp[k++] = s[j];
            }
        }
        temp[k] = '\0';
        
        if (isPalindrome(temp)) {
            free(temp);
            return true;
        }
    }
    
    free(temp);
    return false;
}

int main() {
    char s[100001];
    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(n²)

For each of n positions, we check palindrome in O(n) time

✓ Linear Growth

Space Complexity

O(n)

Creating temporary strings for each deletion attempt

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try All Deletions — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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