This algorithm runs in O(1) space, the O(N) space is required for the output array.

Ya but I was wondering could this be done in-place

this works, without using a 3rd array: def square_arr(arr): i = 0 j = len(arr) - 1

while j > i: sq_j = arr[j] * arr[j] sq_i = arr[i] * arr[i]

if( sq_i > sq_j ) : arr[i] = arr[j] arr[j]= sq_i j -= 1 elif (sq_i < sq_j): arr[j] = sq_j j -= 1 else: j -= 1

print (arr)

  public static int[] makeSquares(int[] arr) {
    int n = arr.length;
    // int[] squares = new int[n];
    // TODO: Write your code here
    int firstPointer = 0, lastPointer = arr.length-1;
    //make a loop here
    while(firstPointer != lastPointer){...