<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> sums;
      int sum = 0;
      for (auto num : nums) {
        sum += num;
        sums.push_back(sum);
      }

      int max_length = nums.size();
      if (nums.size() % 2 != 0) {
        max_length -= 1;
      }

      while(max_length > 0) {
        int l = 0;
        int r = max_length - 1;
        while (r < nums.size()) {
          if (l == 0 && sums[r] == max_length / 2) {
            return max_length;
          } else if(l != 0 && (sums[r] - sums[l-1]) == max_length / 2) {
            return max_length;
          } else {
            l += 1;
            r += 1;
          }
        }
        max_length -= 2;
      }
      return 0;
    }
};
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