Problem of the Day 2/03/2024

My Solution

977. Squares of a Sorted Array

Easy

---

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?

class Solution {
public:
    int getMax(vector<int>& arr) {
        int max = arr[0];
        for (int i = 1; i < arr.size(); i++) {
            if (arr[i] > max) {
                max = arr[i];
            }
        }
        return max;
    }

    void countSort(vector<int>& arr, int exp) {
        vector<int> output(arr.size());
        vector<int> count(10, 0);

        for (int i = 0; i < arr.size(); i++) {
            count[(arr[i] / exp) % 10]++;
        }

        for (int i = 1; i < 10; i++) {
            count[i] += count[i - 1];
        }

        for (int i = arr.size() - 1; i >= 0; i--) {
            output[count[(arr[i] / exp) % 10] - 1] = arr[i];
            count[(arr[i] / exp) % 10]--;
        }

        for (int i = 0; i < arr.size(); i++) {
            arr[i] = output[i];
        }
    }

    void radixSort(vector<int>& arr) {
        int max = getMax(arr);

        for (int exp = 1; max / exp > 0; exp *= 10) {
            countSort(arr, exp);
        }
    }

    vector<int> sortedSquares(vector<int>& nums) {
        for (int i = 0; i < nums.size(); i++) {
            nums[i] = nums[i] * nums[i];
        }
        radixSort(nums);
        return nums;
    }
};