<p>Given a binary array <code>nums</code>, return <em>the maximum length of a contiguous subarray with an equal number of </em><code>0</code><em> and </em><code>1</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [0,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> [0, 1] is the longest contiguous subarray with an equal number of 0 and 1.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,1,0]
<strong>Output:</strong> 2
<strong>Explanation:</strong> [0, 1] (or [1, 0]) is a longest contiguous subarray with equal number of 0 and 1.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>nums[i]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution {
public:
    int findMaxLength(vector<int>& nums) {
        vector<int> positive_sums;
        vector<int> negative_sums;
        positive_sums.push_back(-1);
        negative_sums.push_back(-1);

        int res = 0;
        int sum = 0;
        for(int i = 0; i < nums.size(); ++i)
        {
            sum += (nums[i] == 0)? -1:1;
            int index = -1;
            if(sum > 0) 
            {
                if(positive_sums.size() > sum )
                    index = positive_sums[sum];
                else
                {
                    positive_sums.push_back(i);
                    index = i;
                }
            }
            else
            {
                if(negative_sums.size() > -sum )
                    index = negative_sums[-sum];
                else
                {
                    negative_sums.push_back(i);
                    index = i;
                }
            } 

            res = max(res, i - index);
        }

        return res;
    }
};
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