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Squares of a Sorted Array⚓︎

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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 <= 10^4
  • -10^4 <= nums[i] <= 10^4
  • nums is sorted in non-decreasing order.

Solution⚓︎

Trivial Solution⚓︎

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class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        for (int i = 0; i < nums.size(); i++) nums[i] = nums[i] * nums[i];
        sort(nums.begin(), nums.end());
        return nums;
    }
};

Time complexity: \(O\left( n+n\log n \right) \approx O\left( n\log n \right)\)

Double Pointer Solution⚓︎

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class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        vector<int> res(nums.size(), 0);
        int k = nums.size() - 1;
        for (int i = 0, j = nums.size() - 1; i <= j; ) {  // NOTE!!!  write i <= j because we need to handle the last 2 elements
            if (nums[i] * nums[i] < nums[j] * nums[j]) {
                res[k--] = nums[j] * nums[j];
                j--;
            } else {
                res[k--] = nums[i] * nums[i];
                i++;
            }
        }
        return res;
    }
};

Time complexity: \(O(n)\)