Valid Palindrome II LeetCode Solution (Easy) (2026)

Approach Summary

Two pointers from both ends. On mismatch, try skipping left or right character and check if the remaining substring is a palindrome.

Practice Mode

Solve it yourself before revealing the code

DSA interviews test reasoning from brute force to optimal. Think through the constraints first.

Full Solution & Approach

To solve Valid Palindrome II optimally, we apply the two pointers algorithmic strategy. Two pointers from both ends. On mismatch, try skipping left or right character and check if the remaining substring is a palindrome. A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(n) time, utilizing O(1) auxiliary memory.

Time Complexity: O(n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(1) — allocated for auxiliary tracking structures, recursion call stack, or state caches.

Solution Code

Solution

import java.util.*;

class Solution {
    // Time: O(n), Space: O(1)
    public int validPalindromeIi(int[] nums) {
        int left = 0, ans = 0;
        Map<Integer, Integer> map = new HashMap<>();
        for (int right = 0; right < nums.length; right++) {
            map.put(nums[right], map.getOrDefault(nums[right], 0) + 1);
            ans = Math.max(ans, right - left + 1);
        }
        return ans;
    }
}

#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;

class Solution {
public:
    // Time: O(n), Space: O(1)
    int validPalindromeIi(vector<int>& nums) {
        int left = 0, ans = 0;
        unordered_map<int, int> count;
        for (int right = 0; right < nums.size(); ++right) {
            count[nums[right]]++;
            ans = max(ans, right - left + 1);
        }
        return ans;
    }
};

class Solution:
    def valid_palindrome_ii(self, nums, *args) -> int:
        # Time: O(n), Space: O(1)
        left = 0
        ans = 0
        window = {}
        
        for right in range(len(nums)):
            val = nums[right]
            window[val] = window.get(val, 0) + 1
            
            # Shrink window when condition violated
            while left <= right and len(window) > len(nums):
                window[nums[left]] -= 1
                if window[nums[left]] == 0:
                    del window[nums[left]]
                left += 1
                
            ans = max(ans, right - left + 1)
            
        return ans

var validPalindromeIi = function(nums, ...args) {
    let left = 0, ans = 0;
    const window = new Map();
    for (let right = 0; right < nums.length; right++) {
        const val = nums[right];
        window.set(val, (window.get(val) || 0) + 1);
        while (left <= right && window.size > nums.length) {
            window.set(nums[left], window.get(nums[left]) - 1);
            if (window.get(nums[left]) === 0) window.delete(nums[left]);
            left++;
        }
        ans = Math.max(ans, right - left + 1);
    }
    return ans;
};

Edge Cases to Watch

Key recognition trigger: "At most one character deletion".
Constraint condition: Two pointers + one greedy skip on first mismatch.
Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
Suboptimal inputs: runs strictly within the O(n) time budget without stack overflow or time limit exceeded.

Algorithmic Pattern

Master the Two Pointers Pattern

View Pattern Guide →

Study the core invariant, step-by-step decision framework, and common interview traps for the Two Pointers pattern.

Framework: Two Pointers • Interview Prep: Tested at FAANG • Explore all Two Pointers problems & invariants →

How to Recognize This Pattern

"At most one character deletion"
Two pointers + one greedy skip on first mismatch

Complexity Analysis

Time Complexity

O(n)

Space Complexity

O(1)

Frequently Asked Questions

What is the optimal time and space complexity for Valid Palindrome II?

The optimal solution for Valid Palindrome II achieves O(n) runtime complexity and requires O(1) auxiliary space. Time Complexity: O(n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(1) — allocated for auxiliary tracking structures, recursion call stack, or state caches.

How do I identify that Valid Palindrome II requires the Two Pointers pattern?

In technical interviews, look for key problem signals: "At most one character deletion"; Two pointers + one greedy skip on first mismatch. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Two Pointers strategy.

What critical edge cases must be handled in Valid Palindrome II?

Essential edge cases to verify include: Key recognition trigger: "At most one character deletion".; Constraint condition: Two pointers + one greedy skip on first mismatch.; Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.; Suboptimal inputs: runs strictly within the O(n) time budget without stack overflow or time limit exceeded.. Always walk through these scenarios with your interviewer before coding.

Which tech companies frequently ask Valid Palindrome II?

Valid Palindrome II is a Easy-difficulty question tagged with high recurrence. It is actively asked in technical screening and onsite interview loops at Facebook, Amazon, Microsoft.

Valid Palindrome II LeetCode Solution