977. Squares of a Sorted Array

# O(n) time || O(1) space
def sorted_squares(self, nums: List[int]) -> List[int]:
    res = [0] * len(nums)
    pos = len(nums) - 1
    low, high = 0, len(nums) - 1
    while low <= high:
        if abs(nums[low]) > abs(nums[high]):
            res[pos] = nums[low] ** 2
            low += 1
        else:
            res[pos] = nums[high] ** 2
            high -= 1
        pos -= 1

    return res

Given that the array is sorted in non-decreasing order, the negative numbers will be on the left side and the positive numbers on the right. Squaring the numbers can reverse the order for negative numbers, as larger negative numbers will result in larger squares.

The approach here is to use two pointers technique. One pointer starts at the beginning of the array and the other at the end. We compare the absolute values of the numbers at these pointers, square the larger absolute value, and place it at the current position in our result array (starting from the end). Then, depending on which number we squared, we move the corresponding pointer.