28- Sort Array By Parity

The problem can be found at the following link: Question Link

My Approach

1. Initialize an integer variable j to 0. This variable will keep track of the position where the next even number should be placed in the array.

2. Within the loop, check if the current element at nums[i] is even. You can do this by using the bitwise AND operation with 1, i.e., !(nums[i] & 1). If the result is true, it means the current element is even.

3. Inside the if condition (when the current element is even), perform a swap operation between nums[i] and nums[j]. This swaps the current even element with the element at position j, effectively moving all even elements to the front of the array.

4. Increment the j variable by 1 to prepare for the next even element placement.

5. Continue iterating through the elements of the nums vector, repeating steps 4-6 for each element.

6. Once the loop finishes, the nums vector will be sorted with all even elements at the beginning and all odd elements at the end.

Time and Auxiliary Space Complexity

• Time Complexity: O(n)

• Auxiliary Space Complexity: O(1)

Code (C++)

class Solution {
public:
    vector<int> sortArrayByParity(vector<int>& nums) {
        int j=0;
        for(int i=0;i<nums.size();i++){
            if(!(nums[i]&1)){
                swap(nums[i],nums[j]);
                j++;
            }
        }
        return nums;
    }
};

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