<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:
    vector<int> nums;
    map<pair<int, int>, bool> memo;

    bool canPartition(vector<int>& nums) {

        sort(nums.begin(), nums.end());

        this->nums = nums;

        int sum = 0;
        for (int num : nums) {
            sum += num;
        }

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


        int halfSum = sum / 2;
        return solve(0, halfSum);

        /*
        vector<vector<bool>> dp(nums.size(), vector<bool>(halfSum + 1, false));
        for (int i = 0; i < nums.size(); i++) {
            dp[i][0] = true;
        }

        for (int targetSum = 0; targetSum <= halfSum; targetSum++) {
            for (int i = 0; i < nums.size(); i++) {
                dp[i][targetSum] = targetSum - nums[i] >= 0 ? dp[i][targetSum - nums[i]] : false;
            }
        }
        */
    }

    bool solve(int i, int remainSum) {
        if (remainSum == 0) {
            return true;
        }

        if (i >= nums.size() || remainSum < nums[i]) {
            return false;
        }

        pair<int, int> key = {i, remainSum};
        if (memo.contains(key)) {
            return memo[key];
        }

        bool res = solve(i+1, remainSum - nums[i]);
        res = res || solve(i+1, remainSum);

        return memo[key] = res;
    }
};
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