<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:
        prefix = [0]*len(nums)
        if nums[0] == 0:
            prefix[0] = -1
        else:
            prefix[0] = 1
        for i in range(1,len(nums)):
            if nums[i] == 0:
                prefix[i] = prefix[i-1]+(-1)
            else:
                prefix[i] = prefix[i-1]+nums[i]
        print("prefix array",prefix)
        if prefix[len(nums)-1] == 0:
            print("here")
            return len(nums)
        hash_map = {}
        max_diff = 0
        updated = False
        for i in range(len(prefix)):
            if prefix[i] in hash_map:
                hash_map[prefix[i]].append(i)
                if ((i-hash_map[prefix[i]][0]) > max_diff):
                    max_diff = (i-hash_map[prefix[i]][0])
                    updated = True
            else:
                hash_map[prefix[i]] = [i]
        if 0 in hash_map:
            return max(max_diff,hash_map[0][-1]+1)
        else:
            return max_diff
        
        
        
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