<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>

/*
                        v
    0   1   2   3   4   5   6   7   8
    0   0   1   0   0   0   1   1   0
    
    -1  0   1   4   5   
b = 5
max=2

*/
using iter = vector<int>::iterator;
class Solution {
public:
    int findMaxLength(vector<int>& nums) {
        unordered_map<int, iter> balances;
        balances[0] = prev(nums.begin());

        int balance = 0;
        int maxLength = 0;
        for (iter it = nums.begin(); it != nums.end(); ++it) {
            balance += (*it == 0) ? 1 : -1;
            if (balances.count(balance))
                maxLength = max(maxLength, static_cast<int>(it - balances[balance]));
            else balances[balance] = it;
        }

        return maxLength;
    }
};
89500