3467. Transform Array by Parity

中文文档

Description

You are given an integer array nums. Transform nums by performing the following operations in the exact order specified:

1. Replace each even number with 0.
2. Replace each odd numbers with 1.
3. Sort the modified array in non-decreasing order.

Return the resulting array after performing these operations.

 

Example 1:

Input: nums = [4,3,2,1]

Output: [0,0,1,1]

Explanation:

• Replace the even numbers (4 and 2) with 0 and the odd numbers (3 and 1) with 1. Now, nums = [0, 1, 0, 1].
• After sorting nums in non-descending order, nums = [0, 0, 1, 1].

Example 2:

Input: nums = [1,5,1,4,2]

Output: [0,0,1,1,1]

Explanation:

• Replace the even numbers (4 and 2) with 0 and the odd numbers (1, 5 and 1) with 1. Now, nums = [1, 1, 1, 0, 0].
• After sorting nums in non-descending order, nums = [0, 0, 1, 1, 1].

 

Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 1000

Solutions

Solution 1: Counting

We can traverse the array $\textit{nums}$ and count the number of even elements $\textit{even}$. Then, we set the first $\textit{even}$ elements of the array to $0$ and the remaining elements to $1$.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.

Python3

class Solution:
    def transformArray(self, nums: List[int]) -> List[int]:
        even = sum(x % 2 == 0 for x in nums)
        for i in range(even):
            nums[i] = 0
        for i in range(even, len(nums)):
            nums[i] = 1
        return nums

Java

class Solution {
    public int[] transformArray(int[] nums) {
        int even = 0;
        for (int x : nums) {
            even += (x & 1 ^ 1);
        }
        for (int i = 0; i < even; ++i) {
            nums[i] = 0;
        }
        for (int i = even; i < nums.length; ++i) {
            nums[i] = 1;
        }
        return nums;
    }
}

C++

class Solution {
public:
    vector<int> transformArray(vector<int>& nums) {
        int even = 0;
        for (int x : nums) {
            even += (x & 1 ^ 1);
        }
        for (int i = 0; i < even; ++i) {
            nums[i] = 0;
        }
        for (int i = even; i < nums.size(); ++i) {
            nums[i] = 1;
        }
        return nums;
    }
};

Go

func transformArray(nums []int) []int {
	even := 0
	for _, x := range nums {
		even += x&1 ^ 1
	}
	for i := 0; i < even; i++ {
		nums[i] = 0
	}
	for i := even; i < len(nums); i++ {
		nums[i] = 1
	}
	return nums
}

TypeScript

function transformArray(nums: number[]): number[] {
    const even = nums.filter(x => x % 2 === 0).length;
    for (let i = 0; i < even; ++i) {
        nums[i] = 0;
    }
    for (let i = even; i < nums.length; ++i) {
        nums[i] = 1;
    }
    return nums;
}