<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) {
        int n(nums.size());
        partial_sum(nums.cbegin(), nums.cend(), nums.begin());
        int len(n % 2 == 0 ?  n : n-1);
        while (len > 0) {
            int nOnes = nums[len-1];
            if (nOnes * 2 == len) {
                return len;
            }
            for (int i = 0; i < n - len; i ++) {
                nOnes = nums[i+len] - nums[i];
                if (nOnes * 2 == len) {
                    return len;
                }
            }
            len -= 2;
        }
        return len;
    }
};
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