<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:
        n = len(nums)
        cumsum = [0]*(n+1)
        # find the positions: 
        # j - i = 2*(cumsum[i+1] - cumsum[j+1]), which are count of 1s in between
        max_length = 0
        diff_dict = {0:0}
        for i in range(n):
            cumsum[i+1] = cumsum[i] + nums[i]
            diff = i+1 - 2*cumsum[i+1]
            if diff not in diff_dict:
                diff_dict[diff] = i+1
            else:
                max_length = max(max_length, i+1 - diff_dict[diff])
        return max_length

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