<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:
    def findMaxLength(self, nums: List[int]) -> int:
        from collections import Counter
        counter = Counter(nums)
        if counter[0] == counter[1]:
            return len(nums)

        cum_sum = [0]

        for n in nums:
            if n == 1:
                cum_sum.append(cum_sum[-1] + 1)
            else:
                cum_sum.append(cum_sum[-1] - 1)

        max_len = 0
        d = {}
        for i in range(len(cum_sum)):
            if cum_sum[i] not in d:
                d[cum_sum[i]] = i
            else:
                max_len = max(max_len, i - d[cum_sum[i]])

        return max_len

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