<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:
        ## tackle the cases when half of the elements are 1s
        max_len = len(nums)
        if 2*sum(nums) == max_len:
            return max_len

        nums = [x or -1 for x in nums]
        prefix_sum = 0
        hash_map = {0:-1}
        max_len = 0
        for i, num in enumerate(nums):
            prefix_sum += num           
            if not prefix_sum in hash_map:
                hash_map[prefix_sum] = i
            else:
                max_len = max(max_len, i-hash_map[prefix_sum])
        return max_len
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