<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:
        #equal number between i and j if prefix[j] - prefix[i - 1] = 0 otherwise prefix[j] = prefix[i  - 1] means subarray[i.j] is valid
        #use the map to locate the same and update with current max index.
        m = {1 : 1, 0 : -1}
        res = 0
        prefix = [0] * (len(nums) + 1)
        prefix[0] = m[nums[0]]
        
        for i in range(1,len(nums)):
            prefix[i] = prefix[i-1] + m[nums[i]]
        hm = {}
        for i in range(len(nums)):
            # 0 -> new solution from root
            if prefix[i] == 0:
                res = max(res,i + 1)
            elif prefix[i] in hm:
                res = max(res, i - hm[prefix[i]])
            else:
                hm[prefix[i]] = i
            # in the hm -> new solution from hm index + 1
            #add to hm
        return res
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