<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:
        modified = []
        for n in nums:
            modified.append(-1 if n == 0 else 1)

        prefix_sum = [modified[0]]
        for i in range(1, len(modified)):
            prefix_sum.append(modified[i] + prefix_sum[i-1])

        # If we traverse prefix_sum from the beginning, every 0
        # signifies we've encountered the same number of 1s and 0s since
        # the beginning (i.e. equal number of -1s and +1s).
        #
        # Further, if we encounter the same number at 2 indices i and j,
        # then the sum of the subarray [i, j] is also 0, implying
        # an equal number of 1s and 0s.
        #
        # Track where we encounter each number.

        longest = 0
        seen = {}

        for i, n in enumerate(prefix_sum):
            latest = 0
            if n == 0:
                latest = i + 1
            elif n in seen:
                latest = i - seen[n]
            else:
                seen[n] = i
            
            longest = max(longest, latest)
        
        return longest
23000