<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:
        if nums[0] == 0:
            prefixed_list = [0]
            prefixed_sum = 0
        else:
            prefixed_list = [1]
            prefixed_sum = 1
        number_appeared = 1
        for i in nums:
            if i == 0:
                prefixed_sum -= 1
            else:
                prefixed_sum += 1
            
            prefixed_list.append(prefixed_sum)
            
        count_dict = defaultdict(int)
        maximum_length = 0
        for i in range(len(prefixed_list)):
            if prefixed_list[i] not in count_dict:
                count_dict[prefixed_list[i]] = i
            else:
                maximum_length = max(maximum_length, i - count_dict[prefixed_list[i]])
        return maximum_length
         
        
            
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