<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 {
    struct pair_hash {
    inline std::size_t operator()(const std::pair<int,int> & v) const {
        return v.first*31+v.second;
    }
    };
    using cache = unordered_set<pair<int, int>, pair_hash>;

    bool dfs(int idx, int curr, int target, vector<int>& nums, cache& dp) {
        if (curr == target) {
            return true;
        }

        if (idx >= nums.size() || curr > target) {
            return false;
        }

        if (dp.find(make_pair(idx, curr)) != dp.end()) {
            return false;
        }

        if (dfs(idx+1, curr + nums[idx], target, nums, dp)) {
            return true;
        } 
        if (dfs(idx+1, curr, target, nums, dp)) {
            return true;
        }
        dp.insert(make_pair(idx, curr));
        return false;
        
    }


public:
    bool canPartition(vector<int>& nums) {
        int sum = 0; 
        cache dp;

        for (int i = 0; i < nums.size(); ++i) {
            sum += nums[i];
        }    

        if (sum % 2 != 0) {
            return false;
        }

        int target = sum / 2;
        sum = 0;
        sort(nums.begin(), nums.end());
        return dfs(0, sum, target, nums, dp);
    }
};
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