<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:
        # 000111010 -> -1 -1 -1 1 1 1 -1 1 -1
        # 0: -1
        # 1: -2
        # 2: -3
        # 3: -2
        # 4: -1
        # 5: 0
        # 6: -1
        # 7: 0
        # 8: -1
        nums = [nums[i] if nums[i] == 1 else -1 for i in range(len(nums))]
        sums = []
        last_seen_index = {}
        sum = 0
        for i in range(len(nums)):
            sum += nums[i]
            sums.append(sum)
        print("sums: {}".format(sums))
        max_len = 0
        for i in range(len(sums)):
            local_sum = sums[i]
            # get last seen index of local_sum if exists.
            if local_sum in last_seen_index:
                local_len = i - last_seen_index[local_sum]
                if local_len > max_len:
                    max_len = local_len
            elif local_sum == 0:
                local_len = i + 1
                if local_len > max_len:
                    max_len = local_len
            else:
                last_seen_index[local_sum] = i
        
        return max_len
            
        
23100