<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, arr: List[int]) -> int:

        prefix_sum_map = {}
        prefix_sum_map[0] = -1

        prefix_sum = 0
        max_length = 0

        for j in range(len(arr)):
            if arr[j] == 1:
                prefix_sum += 1
            else:
                prefix_sum -= 1

            if prefix_sum in prefix_sum_map:
                i = prefix_sum_map[prefix_sum]
                max_length = max(max_length, j - i)
            else:
                prefix_sum_map[prefix_sum] = j

        return max_length
   
20000