<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 {
public:
    int findMaxLength(vector<int>& nums) {
        int n = nums.size();
        int N = 2*n+1;
        int arr[N]; 
        fill(arr, arr+N, -2);
        arr[n] = -1;
        
        int cnt = 0, ans = 0;
        for(int i = 0; i<n; i++){
            if(nums[i] == 1) cnt++;
            else cnt--;
            if(arr[cnt+n]>=-1) ans = max(ans, i-arr[cnt+n]);
            else arr[cnt+n] = i;
        }
        
        return ans;
    }
};
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