<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:

    int dp[200][20005];

    bool solve(int i,int sum,int target,vector<int>&nums){
        if(sum==target)return true;
        if(i==nums.size())return false;
        if(sum>target)return false;
        if(dp[i][sum]!=-1)return dp[i][sum];
        int take=solve(i+1,sum+nums[i],target,nums);
        int not_take=solve(i+1,sum,target,nums);
        return dp[i][sum]=take || not_take;
    }

    bool canPartition(vector<int>& nums) {
        int sum=0;
        memset(dp,-1,sizeof(dp));
        for(auto x:nums)sum+=x;
        if(sum%2)return 0;
        sum/=2;
        return solve(0,0,sum,nums);
    }
};
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