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Deque Palindrome Check

Published at25 Jul 2026
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Deque Easy 1 views
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Given an array representing a deque (double-ended queue), determine whether the sequence of elements forms a palindrome — meaning it reads the same forward and backward. Return true if it is a palindrome, false otherwise.

For example, [1, 2, 2, 1] is a palindrome. [1, 2] is not. An empty deque or a single-element deque is always a palindrome. [1, 2, 3, 2, 1] is also a palindrome.

Checking a palindrome using a deque is a classic application of the double-ended queue. Because a deque allows access to both ends, you can compare the front and back elements, removing them simultaneously. If any pair mismatches, it is not a palindrome. This approach is more natural for a deque than using two pointers, because the deque itself manages the endpoints.

In array-based deque implementations, this is equivalent to the two-pointer technique: compare arr[i] with arr[j] where i starts at 0 and j at length-1, moving inward. The deque's ability to pop from both ends makes this operation intuitive and clean.

Edge cases include an empty deque (return true), a single element (return true), even-length palindromes [1, 1] (true), and near-palindromes where only the middle differs [1, 2, 3] (false).

Example Input & Output

Example 1
Input
[]
Output
true
Explanation

Empty

Example 2
Input
[1,2,3,2,1]
Output
true
Explanation

Palindrome

Example 3
Input
[1,2,3,3,2,1]
Output
true
Explanation

Even length palindrome

Example 4
Input
[1,2,3,4]
Output
false
Explanation

Not palindrome

Example 5
Input
[1]
Output
true
Explanation

Single element

Algorithm Flow

Recommendation Algorithm Flow for Deque Palindrome Check

Solution Approach

Use two pointers at both ends to compare elements moving inward.

function solution(arr) {
  for (var i = 0, j = arr.length - 1; i < j; i++, j--) {
    if (arr[i] !== arr[j]) return false;
  }
  return true;
}

Initialize i at the front (0) and j at the back (length-1). Loop while i < j, comparing elements at each position. If any pair differs, return false. If all pairs match, return true. This works for both odd and even length arrays.

Time complexity is O(n) with n being the array length, space complexity is O(1).

Best Answers

java
class Solution {
    public boolean solution(int[] nums) {
        int l=0,r=nums.length-1;while(l<r){if(nums[l]!=nums[r])return false;l++;r--;}return true;
    }
}