<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) {
        unordered_map<int,int> sums;
        int sum=0;
        int ans=0;
        sums[0]=-1;
        for(int i=0;i<nums.size();i++){
            int t=(nums[i]==1?1:-1);
            sum=sum+t;
            if(sums.find(sum)!=sums.end())
            ans=max(ans,i-sums[sum]);
            else
            sums[sum]=i;
        }
        return ans;

        
    }
};
// 1 0 0 0 1 0 0 1 0 0
// 1 0-1-2-1-2
// 0 1 0
// -1 0 1
// 1 0
// 1 0


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