<p>Given an integer array <code>nums</code>, <em>find three numbers whose product is maximum and return the maximum product</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> 6
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [1,2,3,4]
<strong>Output:</strong> 24
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> nums = [-1,-2,-3]
<strong>Output:</strong> -6
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= nums.length &lt;=&nbsp;10<sup>4</sup></code></li>
	<li><code>-1000 &lt;= nums[i] &lt;= 1000</code></li>
</ul>

class Solution {
public:
    int maximumProduct(vector<int>& nums) {
        int first = INT_MIN;
        int second = INT_MIN;
        int third = INT_MIN;

        int firstMin = INT_MAX;
        int secondMin = INT_MAX;
        for (int num: nums) {
            if (num >= first) {
                third = second;
                second = first;
                first = num;
            } else if (num >= second) {
                third = second;
                second = num;
            } else if (num > third) {
                third = num;
            }

            if (num <= firstMin) {
                secondMin = firstMin;
                firstMin = num;
            } else if (num < secondMin) {
                secondMin = num;
            }
        }
        return max(first * second * third, firstMin * secondMin * first);
    }
};
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