Squares of a Sorted Array

LeetCode 1019 | Difficulty: Easy

Easy

Problem Description

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 10^4`

- `-10^4 <= nums[i] <= 10^4`

- `nums` is sorted in **non-decreasing** order.

Follow up: Squaring each element and sorting the new array is very trivial, could you find an O(n) solution using a different approach?

Topics: Array, Two Pointers, Sorting

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Approach

Two Pointers

Use two pointers to traverse the array, reducing the search space at each step. This avoids the need for nested loops, bringing complexity from O(n²) to O(n) or O(n log n) if sorting is involved.

When to use

Array is sorted or can be sorted, and you need to find pairs/triplets that satisfy a condition.

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Solutions

Solution 1: C# (Best: 152 ms)

Metric	Value
Runtime	152 ms
Memory	48.7 MB
Date	2022-01-13

Solution

public class Solution {
    public int[] SortedSquares(int[] nums) {
        int[] result = new int[nums.Length];
        int l = 0, r = nums.Length-1;
        int counter = nums.Length -1;
        while(l <=r)
        {
            if(Math.Abs(nums[l])>Math.Abs(nums[r]))
            {
                result[counter] = nums[l] * nums[l];
                l++;
            }
            else
            {
                result[counter] = nums[r] * nums[r];
                r--;
            }
            counter--;
        }
        return result;
    }
}

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Complexity Analysis

Approach	Time	Space
Two Pointers	$O(n)$	$O(1)$
Sort + Process	$O(n log n)$	$O(1) to O(n)$

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Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• Ask: "Can I sort the array?" — Sorting often enables two-pointer techniques.