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

Solution 1

Python Code
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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

Java Code
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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;
    }
}

C++ Code
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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;
    }
};

Go Code
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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
}

Rust Code
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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
    }
}

JavaScript Code
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/**
 * @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;
};

PHP Code
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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;
    }
}