**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 :

` ````
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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