Square of a List - Amazon Top Interview Questions


Problem Statement :


Given a list of integers sorted in ascending order nums, square the elements and give the output in sorted order.

Constraints

n ≤ 100,000 where n is the length of nums


Example 1

Input
nums = [-9, -2, 0, 2, 3]

Output
[0, 4, 4, 9, 81]



Example 2

Input
nums = [1, 2, 3, 4, 5]

Output
[1, 4, 9, 16, 25]


Solution :



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                        Solution in C++ :

vector<int> solve(vector<int>& nums) {
    int n = nums.size();
    int r = n - 1, l = 0, i = n - 1;
    vector<int> res(n);
    while (l <= r) {
        int rsq = nums[r] * nums[r];
        int lsq = nums[l] * nums[l];
        if (lsq > rsq) {
            res[i] = lsq;
            l++;
        } else {
            res[i] = rsq;
            r--;
        }
        i--;
    }
    return res;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int[] solve(int[] nums) {
        int[] squared = new int[nums.length];
        int i = 0;
        int j = nums.length - 1;
        int k = squared.length - 1;
        while (i <= j) {
            if (nums[i] * nums[i] > nums[j] * nums[j]) {
                squared[k--] = nums[i] * nums[i];
                i++;
            } else {
                squared[k--] = nums[j] * nums[j];
                j--;
            }
        }

        return squared;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums):
        n = len(nums)
        l = 0
        r = n - 1
        index = n - 1
        res = [0 for i in range(len(nums))]
        while index >= 0:
            if abs(nums[l]) > abs(nums[r]):
                res[index] = nums[l] * nums[l]
                l += 1
            else:
                res[index] = nums[r] * nums[r]
                r -= 1
            index -= 1

        return res


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