<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 typing import List
from collections import defaultdict

class Solution:
    def findMaxLength(self, nums: List[int]) -> int:
        # Transform array: convert 0 to -1
        nums = [1 if x == 1 else -1 for x in nums]
        
        # Initialize variables
        max_length = 0
        cumulative_sum = 0
        index_map = {0: -1}  # Handle the case when the subarray starts from index 0
        
        # Iterate over the array
        for i, num in enumerate(nums):
            # Compute the cumulative sum
            cumulative_sum += num
            
            # Check if this cumulative sum has been seen before
            if cumulative_sum in index_map:
                # Calculate the length of the subarray
                length = i - index_map[cumulative_sum]
                max_length = max(max_length, length)
            else:
                # Store the first occurrence of this cumulative sum
                index_map[cumulative_sum] = i
        
        return max_length


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