<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:
        
        count = [0]*(len(nums)+1)        
        
        count_dict = {0:0}
        
        
        total = 0
        max_window_size = 0
        
        
        for idx, num in enumerate(nums):
            
            if num == 1:
                total += 1
            else:
                total -= 1
            
            count[idx+1] = total
            
            if total not in count_dict:
                count_dict[total] = idx+1
            else:
                if idx - count_dict[total] > max_window_size:
                    max_window_size = idx+1 - count_dict[total]
                    
        return max_window_size
            
            
        
            
            
            
            
            
            
            
            
        
                    
                    
                    
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