<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>

from collections import defaultdict
class Solution:
    def findMaxLength(self, nums: List[int]) -> int:
        count_map = defaultdict(int)
        count_map[0] = -1  # To handle cases where subarray starts from index 0
        count = 0
        max_length = 0
        
        for i in range(len(nums)):
            # Convert 0 to -1 to track equal number of 0 and 1
            if nums[i] == 1:
                count += 1
            else:
                count -= 1
            
            if count in count_map:
                # Calculate the length of the subarray
                max_length = max(max_length, i - count_map[count])
            else:
                # Store the first occurrence of the count
                count_map[count] = i
        
        return max_length
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