notes/OJ notes/pages/Leetcode Squares-of-a-Sorted-Array.md

# Leetcode Squares-of-a-Sorted-Array

#### 2022-07-10 08:50

> ##### Algorithms:
> #algorithm #two_pointers 
> ##### Data structures:
> #DS #array 
> ##### Difficulty:
> #coding_problem #difficulty-easy 
> ##### Additional tags:
> #leetcode 
> ##### Revisions:
> N/A

##### Related topics:
##### Links:
- [Link to problem](https://leetcode.com/problems/squares-of-a-sorted-array/)
___
### Problem

Given an integer array `nums` sorted in **non-decreasing** order, return _an array of **the squares of each number** 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?

#### Examples

**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.

### Thoughts

> [!summary]
> This is a #two_pointers 

Using two pointers.
One from left, one from right end.
**fill the array in backwards order.**

### Solution

cpp
```cpp
class Solution {
public:
  vector<int> sortedSquares(vector<int> &nums) {
    // Double pointers, since array is sorted and access in order.
    // O(2n)

    int mid;
    int size = nums.size();
    int left = 0, right = size - 1;
    int count = size - 1;
    vector<int> answer(size);

    for (int i = 0; i < size; i++) {
      nums[i] = nums[i] * nums[i];
    }

    while (left <= right) {
      if (nums[left] >= nums[right]) {
        answer[count--] = nums[left];
        left++;
      } else if (nums[right] > nums[left]) {
        answer[count--] = nums[right];
        right--;
      }
    }

    return answer;
  }
};
```