Valid Palindrome II - Practice Coding Problems

Valid Palindrome II - Problem

Given a string s, determine if it can become a palindrome by deleting at most one character from it.

A palindrome is a string that reads the same forwards and backwards, like "racecar" or "madam". Your task is to check if the given string is already a palindrome, or if removing exactly one character can make it a palindrome.

Examples:

"aba" → true (already a palindrome)
"abca" → true (remove 'c' to get "aba")
"abc" → false (cannot be made palindrome by removing one character)

Input & Output

example_1.py — Basic Case

$ Input: s = "aba"

› Output: true

💡 Note: "aba" is already a palindrome, so no deletion is needed

example_2.py — Single Deletion

$ Input: s = "abca"

› Output: true

💡 Note: We can remove the 'c' at index 2 to get "aba", which is a palindrome

example_3.py — Impossible Case

$ Input: s = "abc"

› Output: false

💡 Note: No single character removal can make "abc" a palindrome. Removing any character gives "bc", "ac", or "ab", none of which are palindromes

Visualization

Tap to expand

Understanding the Visualization

Start Inspection

Place inspectors at both ends of the mirror text

Walk Inward

Move inspectors toward center while characters match

Find Defect

When mismatch found, this is our potential deletion point

Try Fixes

Test removing either left or right character

Verify Fix

Check if remaining text forms perfect palindrome

Key Takeaway

🎯 Key Insight: When two pointers find a mismatch, we only need to try removing one of the two mismatched characters - this eliminates the need to check all possible deletions!

Time & Space Complexity

Time Complexity

⏱️

O(n)

In worst case, we scan the string once and do one additional palindrome check

✓ Linear Growth

Space Complexity

O(1)

Only using a few pointer variables, no extra data structures

✓ Linear Space

Constraints

1 ≤ s.length ≤ 10⁵
s consists of lowercase English letters only
At most one character can be deleted

Asked in

f Facebook 45 ⊞ Microsoft 38 a Amazon 32 G Google 28

The optimal solution uses two pointers starting from both ends. When characters match, move inward. When they don't match, try removing either character and check if the remaining substring is a palindrome. This achieves O(n) time complexity with O(1) space.

Common Approaches

Approach	Time	Space	Notes
✓ Two Pointers (Optimal)	O(n)	O(1)	Use two pointers from ends, when mismatch found, try removing either character
Brute Force (Try All Deletions)	O(n²)	O(n)	Try removing each character one by one and check if result is palindrome

Two Pointers (Optimal) — Algorithm Steps

Initialize left pointer at start (0) and right pointer at end (n-1)
While left < right, compare characters at left and right positions
If characters match, move both pointers inward (left++, right--)
If characters don't match, we have two options:
- Skip left character and check if substring[left+1...right] is palindrome
- Skip right character and check if substring[left...right-1] is palindrome
If either option results in a palindrome, return true
If neither option works, return false

Visualization

Tap to expand

Step-by-Step Walkthrough

Initialize

Set left=0, right=3 for string 'abca'

First Match

s[0]='a' == s[3]='a', move pointers inward

Mismatch Found

s[1]='b' != s[2]='c', try both removal options

Check Options

Option 1: remove 'b' → check 'ca'. Option 2: remove 'c' → check 'ba'

Code -

solution.c — C

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

bool isPalindromeRange(char* s, int left, int right) {
    while (left < right) {
        if (s[left] != s[right]) {
            return false;
        }
        left++;
        right--;
    }
    return true;
}

bool validPalindrome(char* s) {
    int left = 0, right = strlen(s) - 1;
    
    while (left < right) {
        if (s[left] != s[right]) {
            return isPalindromeRange(s, left + 1, right) || 
                   isPalindromeRange(s, left, right - 1);
        }
        left++;
        right--;
    }
    
    return true;
}

int main() {
    char s[1001];
    scanf("%s", s);
    printf("%s\n", validPalindrome(s) ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

In worst case, we scan the string once and do one additional palindrome check

✓ Linear Growth

Space Complexity

O(1)

Only using a few pointer variables, no extra data structures

✓ Linear Space

Constraints

1 ≤ s.length ≤ 10⁵
s consists of lowercase English letters only
At most one character can be deleted

68.2K Views

High Frequency

~15 min Avg. Time

1.8K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Two Pointers (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

Select Compiler