< Back 977. Squares of a Sorted Array This is another pattern of the two-pointer problem. One could trivially solve this problem in O(n log n) time by squaring all of the values and sorting the array, however, we can do better and achieve a solution in O(n) time. We maintain two pointers - one at the beginning of the array and one at the end. We also create another array to store the result as we won't be able to solve this problem in-place. Working backwards from the end of the array, we compare the absolute values of the integers at each pointer. If the value pointed to by the left pointer is greater, we add it to the end of the results array and square the value. We then increment the left pointer. We do the same for the right pointer, however, we decrement the right pointer instead. Eventually we evaluate all of the integers in the array and return the result. The solution is as follows: class Solution: def sortedSquares(self, nums: List[int]) -> List[int]: n = len(nums) result = [0] * n l, r = 0, n - 1 for i in range(n - 1, -1, -1): if abs(nums[l]) < abs(nums[r]): result[i] = nums[r] r -= 1 else: result[i] = nums[l] l += 1 result[i] *= result[i] return result _ Time Complexity: O(n) - We iterate through the entire array once. _ Space Complexity: O(n) - We create a new array to store the result.