<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][40001];
    bool isPossible(vector<int> &nums, int i,int sum, int target){
        if(target==sum){
            return true;
        }
        if(i >= nums.size()){
            return false;
        }

        if(dp[i][sum]!=-1) return dp[i][sum];

        bool taken = isPossible(nums, i + 1, sum + nums[i], target);
        bool notTaken = isPossible(nums, i + 1, sum, target);
        bool res = taken or notTaken;
        return dp[i][sum] = res;
    }
    bool canPartition(vector<int>& nums) {
        int sum = accumulate(nums.begin(), nums.end(), 0);
        if(sum&1) return false;
        memset(dp, -1, sizeof(dp));
        return isPossible(nums, 0, 0, sum/2);
    }
};
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