<p>Given an integer array <code>nums</code>, move all the even integers at the beginning of the array followed by all the odd integers.</p>

<p>Return <em><strong>any array</strong> that satisfies this condition</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0]
<strong>Output:</strong> [0]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 5000</code></li>
</ul>

class Solution {
    // Going to try and use the bitwise (nums[i] & 1) to be quicker than
    // nums[i] % 2
public:
    vector<int> sortArrayByParity(vector<int>& nums) {
        int i = 0;
        int j = nums.size() - 1;
        // int temp = 0;

        while (i <= j) {
            if (((nums[i] & 1) == 1) & ((nums[j] & 1) == 0)) {
                // If start is odd and end is even, swap
                /*
                temp = nums[i];
                nums[i] = nums[j];
                nums[j] = temp;
                */
                // temp variable is not the most effiecient, going to try XOR

                nums[i] = nums[i] ^ nums[j];
                nums[j] = nums[i] ^ nums[j];
                nums[i] = nums[i] ^ nums[j];

                i++;
                j--;
            } else if (((nums[i] & 1) == 1) & ((nums[j] & 1) == 1)) {
                // if both are odd, move end pointer down to try and find an
                // even
                j--;
            } else if (((nums[i] & 1) == 0) & ((nums[j] & 1) == 0)) {
                // if both are even, move start pointer up to try and find an
                // odd
                i++;
            } else {
                // if both are correct, move both towards the center
                i++;
                j--;
            }
        }

        return nums;
    }
};
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