<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 dp[200][100000];
class Solution {
public:
    bool util(vector<int>& nums, int index, int n, int sum, int curr) {
        if (curr == sum - curr) return 1;
        if (index == n) return 0;
        if (dp[index][curr] != -1) return dp[index][curr];
        bool include = util(nums, index + 1, n, sum, curr + nums[index]);
        bool exclude = util(nums, index + 1, n, sum, curr);
        return dp[index][curr] = include || exclude;
    }
    bool canPartition(vector<int>& nums) {
        int n = nums.size(), sum = 0;
        for (int i = 0; i < n; i++) {
            sum += nums[i];
        }
        memset(dp, -1, sizeof(dp));
        return util(nums, 0, n, sum, 0);
    }
};
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