Sort Array by Parity II

Problem Description

Given an array of integers nums, half of the integers in nums are odd, and the other half are even.

Sort the array so that whenever nums[i] is odd, i is odd, and whenever nums[i] is even, i is even.

Return any answer array that satisfies this condition.

Examples

Example 1:

Input: nums = [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.

Example 2:

Input: nums = [2,3]
Output: [2,3]

Constraints

2 <= nums.length <= 2 * 10^4
nums.length is even.
Half of the integers in nums are even.
0 <= nums[i] <= 1000

Solution

Python

class Solution:
    def sortArrayByParityII(self, nums: List[int]) -> List[int]:
        even_index, odd_index = 0, 1
        while even_index < len(nums) and odd_index < len(nums):
            if nums[even_index] % 2 == 0:
                even_index += 2
            elif nums[odd_index] % 2 == 1:
                odd_index += 2
            else:
                nums[even_index], nums[odd_index] = nums[odd_index], nums[even_index]
                even_index += 2
                odd_index += 2
        return nums

# Example usage:
solution = Solution()
print(solution.sortArrayByParityII([4,2,5,7]))  # Output: [4,5,2,7]

C++

#include <vector>
using namespace std;

class Solution {
public:
    vector<int> sortArrayByParityII(vector<int>& nums) {
        int even_index = 0, odd_index = 1;
        while (even_index < nums.size() && odd_index < nums.size()) {
            if (nums[even_index] % 2 == 0) {
                even_index += 2;
            } else if (nums[odd_index] % 2 == 1) {
                odd_index += 2;
            } else {
                swap(nums[even_index], nums[odd_index]);
                even_index += 2;
                odd_index += 2;
            }
        }
        return nums;
    }
};

// Example usage:
int main() {
    Solution solution;
    vector<int> nums = {4, 2, 5, 7};
    vector<int> result = solution.sortArrayByParityII(nums);  // Output: [4, 5, 2, 7]
    for (int num : result) {
        cout << num << " ";
    }
    return 0;
}

Java

class Solution {
    public int[] sortArrayByParityII(int[] nums) {
        int evenIndex = 0, oddIndex = 1;
        while (evenIndex < nums.length && oddIndex < nums.length) {
            if (nums[evenIndex] % 2 == 0) {
                evenIndex += 2;
            } else if (nums[oddIndex] % 2 == 1) {
                oddIndex += 2;
            } else {
                int temp = nums[evenIndex];
                nums[evenIndex] = nums[oddIndex];
                nums[oddIndex] = temp;
                evenIndex += 2;
                oddIndex += 2;
            }
        }
        return nums;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        int[] nums = {4, 2, 5, 7};
        int[] result = solution.sortArrayByParityII(nums);  // Output: [4, 5, 2, 7]
        for (int num : result) {
            System.out.print(num + " ");
        }
    }
}

JavaScript

var sortArrayByParityII = function (nums) {
  let evenIndex = 0,
    oddIndex = 1;
  while (evenIndex < nums.length && oddIndex < nums.length) {
    if (nums[evenIndex] % 2 === 0) {
      evenIndex += 2;
    } else if (nums[oddIndex] % 2 === 1) {
      oddIndex += 2;
    } else {
      [nums[evenIndex], nums[oddIndex]] = [nums[oddIndex], nums[evenIndex]];
      evenIndex += 2;
      oddIndex += 2;
    }
  }
  return nums;
};

// Example usage:
console.log(sortArrayByParityII([4, 2, 5, 7])); // Output: [4, 5, 2, 7]