<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>

//bit manipulation, sliding window, dp, prefix
class Solution {
public:
    int findMaxLength(vector<int>& nums) {
        int n = nums.size();
        unordered_map<int, pair<int, int>> m;

        for(int i=0; i<n; i++){
            if(nums[i]==0) nums[i]=-1;
        }

        m[0].first = -1;
        m[0].second = -1;

        int sum =0;
        for(int i=0; i<n; i++){
            sum+= nums[i];
            if(m.find(sum)==m.end()) m[sum].first = i;
            else m[sum].second = i;
        }

        int res=0;
        for(auto a:m){
            res = max(res, a.second.second - a.second.first);   
        }

        return res;
    }
};
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