Squares of a Sorted Array

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 <= nums.length <= 104
• -104 <= nums[i] <= 104
• 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?

Code

$O(n\log n)$ time, $O(1)$ space

class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        for(auto& n: nums) {
            n = n * n;
        }
        sort(nums.begin(), nums.end());
        return nums;
    }
};

$O(n)$ time, $O(n)$ space Using two pointer

class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        vector<int> answer(nums.size());
        int l = 0, r = nums.size() - 1;
        for(int k = nums.size() - 1; k >= 0; k--) {
            if(abs(nums[l]) > abs(nums[r])) {
                answer[k] = pow(nums[l++], 2);
            } else {
                answer[k] = pow(nums[r--], 2);     
            }
        }
        return answer;
    }
};

Source

• Squares of a Sorted Array - LeetCode