<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 {
    // for other solution of this u can refer to sam equestion in gfg.
    private:
     // optimizing space
    bool solveSO(int N, vector<int>& nums, int total){
       
       vector<int> curr(total+1, 0);
       vector<int> next(total+1, 0);
       
       curr[0] = 1;
       next[0] = 1;
        
        for(int index = N-1; index>=0; index--){
            for(int target = 1; target <= total/2; target++){
                
                bool incl = 0;
                if(target - nums[index] >= 0){
                    
                    incl = next[target-nums[index]];
                    
                }
                bool excl = next[target-0];
                
                curr[target] = (incl or excl);
            }
            next = curr;
        }
        
        return next[total/2] ;
    }
public:
    bool canPartition(vector<int>& nums) {
        int N = nums.size();
        int total = 0;
        for(int i = 0; i<N; i++){
            total += nums[i];
        }
        
        if(total & 1){
            // means odd
            return 0;
        }
        
        int target = total/2;
        
        return solveSO(N, nums, total);
    }
};
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