0977 - Squares of A Sorted Array (Easy)

Problem Link

https://leetcode.com/problems/squares-of-a-sorted-array/

Problem Statement

Given an integer array nums sorted in non-decreasing order, return an array of the squares of each number sorted in non-decreasing order.

Example 1:

Input: nums = [-4,-1,0,3,10]
Output: [0,1,9,16,100]
Explanation: After squaring, the array becomes [16,1,0,9,100]. After sorting, it becomes [0,1,9,16,100]. 

Example 2:

Input: nums = [-7,-3,2,3,11]
Output: [4,9,9,49,121]

Constraints:

• $1 <= key.length, value.length <= 10^4$
• $-10^4 <= nums[i] <= 10^4$
• nums is sorted in non-decreasing order.

Approach 1: Two Pointers

Because the input array is sorted but may include negative numbers, the largest square value may come from either end of the array (a large negative or a large positive number). To handle this, use two pointers l and r starting at each end of the array. Compare abs(nums[l]) and abs(nums[r]). The larger absolute value has the larger square.

Place that square at position pos (starting from the end of the result array), then move the corresponding pointer inward (l += 1 or r -= 1). Decrement pos each step and repeat until all elements are processed.

Time Complexity: $O(n)$

Space Complexity: $O(n)$

class Solution:
    def sortedSquares(self, nums: List[int]) -> List[int]:
        n = len(nums)
        res = [0] * n
        l, r = 0, n - 1
        pos = n - 1
        while l <= r:
            if abs(nums[l]) < abs(nums[r]):
                res[pos] = nums[r] ** 2
                r -= 1
            else:
                res[pos] = nums[l] ** 2
                l += 1
            pos -= 1
        return res