<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[201][20001];

    bool compute(int n, int t, vector<int> &nums) {
        if(n == 0) return false;
        if(t == 0) return true;

        if(dp[n][t] != -1) return dp[n][t];

        if(t-nums[n-1] >= 0)
            return dp[n][t] = compute(n-1, t-nums[n-1], nums) or compute(n-1, t, nums);
        else
            return dp[n][t] = compute(n-1, t, nums);
    }

    bool canPartition(vector<int>& nums) {
        memset(dp, -1, sizeof(dp));

        int sum = 0;
        int n = nums.size();

        for(int i=0; i<n; i++) {
            sum += nums[i];
        }
        if(sum%2) return false;

        return compute(n, sum/2, nums);
    }
};
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