Welcome to Subscribe On Youtube

3467. Transform Array by Parity

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 $\mathit{\text{nums}}$ and count the number of even elements $\mathit{\text{even}}$. Then, we set the first $\mathit{\text{even}}$ 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 $\mathit{\text{nums}}$. The space complexity is $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  
  

• 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;
      }
  };
  
  

• 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
  
  

• 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
  }
  
  

• 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;
  }
  
  

All Problems

All Solutions