<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(); 
        vector<int> sums(n, 0);
        unordered_map<int, int> hm;
        sums[0] = (nums[0] ? 1 : -1);
        for(int i = 1; i < n; i++)
            sums[i] = sums[i-1] + (nums[i] ? 1 : -1);
        if(sums[n-1] == 0)
            return n;
        int mx = 0;
        for(int i = 0; i < n; i++) {
            if(sums[i] == 0)
                mx = i+1;
            if(hm.count(sums[i]))
                mx = max(mx, i - hm[sums[i]]);
            else
                hm[sums[i]] = i; 
        } 
        return mx;
    }
};
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