<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of <strong>even</strong> length consisting of an <strong>equal</strong> number of positive and negative integers.</p>

<p>You should return the array of nums such that the the array follows the given conditions:</p>

<ol>
	<li>Every <strong>consecutive pair</strong> of integers have <strong>opposite signs</strong>.</li>
	<li>For all integers with the same sign, the <strong>order</strong> in which they were present in <code>nums</code> is <strong>preserved</strong>.</li>
	<li>The rearranged array begins with a positive integer.</li>
</ol>

<p>Return <em>the modified array after rearranging the elements to satisfy the aforementioned conditions</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [3,1,-2,-5,2,-4]
<strong>Output:</strong> [3,-2,1,-5,2,-4]
<strong>Explanation:</strong>
The positive integers in nums are [3,1,2]. The negative integers are [-2,-5,-4].
The only possible way to rearrange them such that they satisfy all conditions is [3,-2,1,-5,2,-4].
Other ways such as [1,-2,2,-5,3,-4], [3,1,2,-2,-5,-4], [-2,3,-5,1,-4,2] are incorrect because they do not satisfy one or more conditions.  
</pre>

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

<pre>
<strong>Input:</strong> nums = [-1,1]
<strong>Output:</strong> [1,-1]
<strong>Explanation:</strong>
1 is the only positive integer and -1 the only negative integer in nums.
So nums is rearranged to [1,-1].
</pre>

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

<ul>
	<li><code>2 &lt;= nums.length &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>nums.length</code> is <strong>even</strong></li>
	<li><code>1 &lt;= |nums[i]| &lt;= 10<sup>5</sup></code></li>
	<li><code>nums</code> consists of <strong>equal</strong> number of positive and negative integers.</li>
</ul>

<p>&nbsp;</p>
It is not required to do the modifications in-place.
class Solution {
public:
    vector<int> rearrangeArray(vector<int>& nums) {
        int neg = 0;
        int pos = 0;
        int i = 0;
        int n = nums.size();

        // Find the first negative and positive indices
        while (nums[neg] >= 0) neg++;
        while (nums[pos] < 0) pos++;

        // Alternately push positive and negative numbers to the back
        for (int i = 0; i < n; i++) {
            if (i % 2 == 0) {
                nums.push_back(nums[pos]);
                pos++;
                while (pos < n && nums[pos] < 0) pos++;  // Find the next positive number
            } else {
                nums.push_back(nums[neg]);
                neg++;
                while (neg < n && nums[neg] >= 0) neg++;  // Find the next negative number
            }
        }

        // Erase the original first half of the array
        nums.erase(nums.begin(), nums.begin() + n);

        return nums;
    }
};

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