### An algorithm to check if a given array is palindrome or not.

We will use two different methods to check if a given array is a palindrome or not in javascript. Everything will be written in ES6.

**Palindrome**: A word, sequence or number that reads same when reversed.

#### Example

Input: [1, 2, 3, 2, 1] [1, 2, 3, 3, 1] Output: true false

### Brute force method

#### Implementation

- We are going to loop through the half of the array.
- And we will check if first half of the array is equal to the other half or not.
- If it is equal then return
`true`

else return`false`

.

let palindromeArray = (arr) => { //initialize to true let isPalindrome = true; //loop through half length of the array for(let i = 0; i < arr.length / 2; i++) { //check if first half is equal to the second half if(arr[i] !== arr[arr.length - i - 1]){ isPalindrome = false; break; } } return isPalindrome; }

Input: console.log(palindromeArray([1,2,2,1])); console.log(palindromeArray([1,2,2,2])); Output: true false

Time complexity: O(n).

Space complexity: O(1).

#### Time and Space complexity

- We are looping through half of the array O(n/2), so Time complexity is O(n).
- We are using constant space, so Space complexity is O(1).

### Using recursion to check the palindrome array.

We can use recursive functions as well to check the palindrome array in javascript.

#### Implementation

- We will create a function and check if first element of the given array is equal to the last element then call the function again with second element and second last element. Repeat this till there is only 1 element.
- If array contains only one element then return
`true`

else return`false`

.

let palindromeArray = (arr, start = 0, end = arr.length - 1) => { //if array has only element if(start >= end){ return true; } //if start is equal to end if(arr[start] === arr[end]){ //call the same function return palindromeArray(arr, start + 1, end - 1); }else{ //else not palindrome return false; } }

Input: console.log(palindromeArray([1,2,2,1])); console.log(palindromeArray([1,2,2,2])); Output: true false

We are using higher-order functions of ES6 in which we can assign the left parameter value to the right parameters `(arr, start = 0, end = arr.length - 1)`

. Learn more about it here.

Time complexity: O(n).

Space complexity: O(n).

#### Time and Space complexity

- We are calling the same function for half of the array O(n/2). so Time complexity is O(n).
- We are storing each function in call stack, so Space complexity is O(n).