<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:
        nums_len = len(nums)
        if 2*sum(nums) == nums_len:
            return nums_len
        
        hmap = {}
        ans = 0
        psum = 0
        for i,v in enumerate(nums):
            psum += 1 if v==1 else -1
            if psum == 0:
                ans = i + 1 # i is 0 indexed
            else:
                if psum in hmap:
                    ans = max(ans, i - hmap[psum])
                else:
                    hmap[psum] = i
        return ans
18400