<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> p(n,1);
        for(int i=0;i<n;i++) if(nums[i]==0) nums[i]=-1;
        p[0]=nums[0];
        for(int i=1;i<n;i++) p[i]=p[i-1]+nums[i];
        // for(auto i:p) cout<<i<<" ";
        int ans=0;
        unordered_map<int,int> mp;
        mp[0]=-1;
        for(int i=0;i<n;i++){
            if(mp.find(p[i])!=mp.end()){
                ans=max(ans,i-mp[p[i]]);
            }
            else{
                mp[p[i]]=i;
            }
        }
        return ans;
    }
};
91500