<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):
        arr = [-2] * (2 * len(nums) + 1)
        arr[len(nums)] = -1
        maxlen = 0
        count = 0
        for i in range(len(nums)):
            count += -1 if nums[i] == 0 else 1
            if arr[count + len(nums)] >= -1:
                maxlen = max(maxlen, i - arr[count + len(nums)])
            else:
                arr[count + len(nums)] = i
        return maxlen


class Solution0:
    def findMaxLength(self, nums: List[int]) -> int:
        
        opt = 2 * min(sum(nums), len(nums)-sum(nums))
        hlfopt = opt // 2

        runsum = [0] * (len(nums) + 1)
        for i in range(1, len(nums)+1):
            runsum[i] = sum(nums[: i])

        while opt > 1:
            for i in range(0, len(nums)-opt+1):
                # print(sum(nums[i: i+opt]), runsum[i+opt]-runsum[i])
                # if sum(nums[i: i+opt]) == hlfopt:
                if runsum[i+opt]-runsum[i] == hlfopt:
                    return opt
            opt -= 2
            hlfopt -= 1
        
        return 0
            
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