<p>Given an integer array <code>nums</code>, return <code>true</code> <em>if you can partition the array into two subsets such that the sum of the elements in both subsets is equal or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,5,11,5]
<strong>Output:</strong> true
<strong>Explanation:</strong> The array can be partitioned as [1, 5, 5] and [11].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,5]
<strong>Output:</strong> false
<strong>Explanation:</strong> The array cannot be partitioned into equal sum subsets.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 200</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
</ul>

class Solution {
public:
    bool canPartition(vector<int>& a) {
        int n=a.size();
        int s=0;
        for(auto i:a)
            s+=i;
        if(s%2)
            return false;
        int x=0,sz;
        queue<int> q;
        unordered_set<int> visited;
        q.push(0);
        for(int i=n-1;i>=0;i--){
            sz=q.size();
            while(sz--){
                x=q.front();
                q.pop();
                if(x==s/2 || x+a[i]==s/2)
                    return true;
                if(x+a[i]<s/2 && !visited.count(x+a[i])){
                    visited.insert(x+a[i]);
                    q.push(x+a[i]);
                }
                q.push(x);
            }
        }
        return false;
    }

};
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