<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>

int t[201][100005];
class Solution {
public:
    bool isSubsetPos(vector<int>& nums,int n,int sum){
        if(n==0){
            return false;
        }
        if(sum==0){
            return true;
        }
        if(t[n][sum]!=-1){
            return t[n][sum];
        }
        if(nums[n-1]<=sum){
            return t[n][sum]=isSubsetPos(nums,n-1,sum-nums[n-1]) || isSubsetPos(nums,n-1,sum);
        }else{
            return t[n][sum]=isSubsetPos(nums,n-1,sum);
        }
    }
    bool canPartition(vector<int>& nums) {
        int sum=0;
        int n=nums.size();
        memset(t,-1,sizeof(t));
        for(int i=0;i<nums.size();i++){
          sum=sum+nums[i];
        }
        if(sum%2==0){
            return isSubsetPos(nums,n,sum/2);
        }
        else{
            return false;
        }
    }
};
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