<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) {

        if(nums.size()<2)
        return 0;

        //find the sum of all elements in the array

        int sum=0, lsum=0, rsum=0, tmp=0;

        for(int i:nums){
            sum+=i;
        }

         
        int wnd=nums.size();

        if((wnd%2)!=0){
        wnd--;
        rsum=nums[nums.size()-1];
        tmp=rsum;
        }

        while(wnd>=2){
            
            for(int st=0; st<=nums.size()-wnd; st++){

                if((sum-(lsum+rsum))==(wnd/2))
                return wnd;
                
            
            lsum+=nums[st];
            if((st+wnd)<nums.size())
            rsum-=nums[st+wnd];
            }

            lsum=0;
             tmp+=(nums[wnd-1]+nums[wnd-2]);
            rsum=tmp;

        wnd-=2;
        }

    return 0;
        
    }
};
82800