905. Sort Array By Parity

Problem Description

Given an integer array nums, move all the even integers at the beginning of the array followed by all the odd integers.

Return any array that satisfies this condition.

Example 1:

Input: nums = [3,1,2,4]
Output: [2,4,3,1]
Explanation: The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.

Example 2:

Input: nums = [0]
Output: [0]

Constraints:

• 1 <= nums.length <= 5000
• 0 <= nums[i] <= 5000

Python Solution

class Solution:
    def sortArrayByParity(self, nums: List[int]) -> List[int]:
        left, right = 0, len(nums) - 1
        
        while left < right:
            if nums[left] % 2 == 1 and nums[right] % 2 == 0:
                nums[left], nums[right] = nums[right], nums[left]
            
            if nums[left] % 2 == 0:
                left += 1
            if nums[right] % 2 == 1:
                right -= 1
        
        return nums

Java Solution

class Solution {
    public int[] sortArrayByParity(int[] nums) {
        int left = 0, right = nums.length - 1;
        
        while (left < right) {
            if (nums[left] % 2 == 1 && nums[right] % 2 == 0) {
                int temp = nums[left];
                nums[left] = nums[right];
                nums[right] = temp;
            }
            
            if (nums[left] % 2 == 0) {
                left++;
            }
            if (nums[right] % 2 == 1) {
                right--;
            }
        }
        
        return nums;
    }
}

C++ Solution

class Solution {
public:
    vector sortArrayByParity(vector& nums) {
        int left = 0, right = nums.size() - 1;
        
        while (left < right) {
            if (nums[left] % 2 == 1 && nums[right] % 2 == 0) {
                swap(nums[left], nums[right]);
            }
            
            if (nums[left] % 2 == 0) {
                left++;
            }
            if (nums[right] % 2 == 1) {
                right--;
            }
        }
        
        return nums;
    }
};

JavaScript Solution

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var sortArrayByParity = function(nums) {
    let left = 0, right = nums.length - 1;
    
    while (left < right) {
        if (nums[left] % 2 === 1 && nums[right] % 2 === 0) {
            [nums[left], nums[right]] = [nums[right], nums[left]];
        }
        
        if (nums[left] % 2 === 0) {
            left++;
        }
        if (nums[right] % 2 === 1) {
            right--;
        }
    }
    
    return nums;
};

C# Solution

public class Solution {
    public int[] SortArrayByParity(int[] nums) {
        int left = 0, right = nums.Length - 1;
        
        while (left < right) {
            if (nums[left] % 2 == 1 && nums[right] % 2 == 0) {
                int temp = nums[left];
                nums[left] = nums[right];
                nums[right] = temp;
            }
            
            if (nums[left] % 2 == 0) {
                left++;
            }
            if (nums[right] % 2 == 1) {
                right--;
            }
        }
        
        return nums;
    }
}