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977. Squares of a Sorted Array

Jump to Solution: Python Java C++ JavaScript C#

Problem 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 <= 10^4
-10^4 <= nums[i] <= 10^4
nums is sorted in non-decreasing order.

Solution

Python Solution

class Solution:
    def sortedSquares(self, nums: List[int]) -> List[int]:
        n = len(nums)
        result = [0] * n
        left, right = 0, n - 1
        
        for i in range(n - 1, -1, -1):
            if abs(nums[left]) > abs(nums[right]):
                result[i] = nums[left] * nums[left]
                left += 1
            else:
                result[i] = nums[right] * nums[right]
                right -= 1
                
        return result

Time Complexity: O(n)

We need to process each element exactly once.

Space Complexity: O(1)

We only use a constant amount of extra space.

Java Solution

class Solution {
    public int[] sortedSquares(int[] nums) {
        int n = nums.length;
        int[] result = new int[n];
        int left = 0, right = n - 1;
        
        for (int i = n - 1; i >= 0; i--) {
            if (Math.abs(nums[left]) > Math.abs(nums[right])) {
                result[i] = nums[left] * nums[left];
                left++;
            } else {
                result[i] = nums[right] * nums[right];
                right--;
            }
        }
        
        return result;
    }
}

Time Complexity: O(n)

We need to process each element exactly once.

Space Complexity: O(1)

We only use a constant amount of extra space.

C++ Solution

class Solution {
public:
    vector sortedSquares(vector& nums) {
        int n = nums.size();
        vector result(n);
        int left = 0, right = n - 1;
        
        for (int i = n - 1; i >= 0; i--) {
            if (abs(nums[left]) > abs(nums[right])) {
                result[i] = nums[left] * nums[left];
                left++;
            } else {
                result[i] = nums[right] * nums[right];
                right--;
            }
        }
        
        return result;
    }
};

Time Complexity: O(n)

We need to process each element exactly once.

Space Complexity: O(1)

We only use a constant amount of extra space.

JavaScript Solution

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var sortedSquares = function(nums) {
    const n = nums.length;
    const result = new Array(n);
    let left = 0, right = n - 1;
    
    for (let i = n - 1; i >= 0; i--) {
        if (Math.abs(nums[left]) > Math.abs(nums[right])) {
            result[i] = nums[left] * nums[left];
            left++;
        } else {
            result[i] = nums[right] * nums[right];
            right--;
        }
    }
    
    return result;
};

Time Complexity: O(n)

We need to process each element exactly once.

Space Complexity: O(1)

We only use a constant amount of extra space.

C# Solution

public class Solution {
    public int[] SortedSquares(int[] nums) {
        int n = nums.Length;
        int[] result = new int[n];
        int left = 0, right = n - 1;
        
        for (int i = n - 1; i >= 0; i--) {
            if (Math.Abs(nums[left]) > Math.Abs(nums[right])) {
                result[i] = nums[left] * nums[left];
                left++;
            } else {
                result[i] = nums[right] * nums[right];
                right--;
            }
        }
        
        return result;
    }
}

Time Complexity: O(n)

We need to process each element exactly once.

Space Complexity: O(1)

We only use a constant amount of extra space.