# 977. Squares of a Sorted Array

## 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 <= 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?

## Solutions

• class Solution {
public int[] sortedSquares(int[] nums) {
int n = nums.length;
int[] res = new int[n];
for (int i = 0, j = n - 1, k = n - 1; i <= j;) {
if (nums[i] * nums[i] > nums[j] * nums[j]) {
res[k--] = nums[i] * nums[i];
++i;
} else {
res[k--] = nums[j] * nums[j];
--j;
}
}
return res;
}
}

• class Solution {
public:
vector<int> sortedSquares(vector<int>& nums) {
int n = nums.size();
vector<int> res(n);
for (int i = 0, j = n - 1, k = n - 1; i <= j;) {
if (nums[i] * nums[i] > nums[j] * nums[j]) {
res[k--] = nums[i] * nums[i];
++i;
} else {
res[k--] = nums[j] * nums[j];
--j;
}
}
return res;
}
};

• class Solution:
def sortedSquares(self, nums: List[int]) -> List[int]:
n = len(nums)
res = [0] * n
i, j, k = 0, n - 1, n - 1
while i <= j:
if nums[i] * nums[i] > nums[j] * nums[j]:
res[k] = nums[i] * nums[i]
i += 1
else:
res[k] = nums[j] * nums[j]
j -= 1
k -= 1
return res


• func sortedSquares(nums []int) []int {
n := len(nums)
res := make([]int, n)
for i, j, k := 0, n-1, n-1; i <= j; {
if nums[i]*nums[i] > nums[j]*nums[j] {
res[k] = nums[i] * nums[i]
i++
} else {
res[k] = nums[j] * nums[j]
j--
}
k--
}
return res
}

• /**
* @param {number[]} nums
* @return {number[]}
*/
var sortedSquares = function (nums) {
const n = nums.length;
const res = new Array(n);
for (let i = 0, j = n - 1, k = n - 1; i <= j; ) {
if (nums[i] * nums[i] > nums[j] * nums[j]) {
res[k--] = nums[i] * nums[i];
++i;
} else {
res[k--] = nums[j] * nums[j];
--j;
}
}
return res;
};


• class Solution {
/**
* @param Integer[] $nums * @return Integer[] */ function sortedSquares($nums) {
$i = 0;$j = $k = count($nums) - 1;
$rs = array_fill(0, count($nums), -1);
while ($i <=$j) {
$max1 =$nums[$i] *$nums[$i];$max2 = $nums[$j] * $nums[$j];
if ($max1 >$max2) {
$rs[$k] = $max1;$i++;
} else {
$rs[$k] = $max2;$j--;
}
$k--; } return$rs;
}
}

• impl Solution {
pub fn sorted_squares(nums: Vec<i32>) -> Vec<i32> {
let n = nums.len();
let mut l = 0;
let mut r = n - 1;
let mut res = vec![0; n];
for i in (0..n).rev() {
let a = nums[l] * nums[l];
let b = nums[r] * nums[r];
if a < b {
res[i] = b;
r -= 1;
} else {
res[i] = a;
l += 1;
}
}
res
}
}