<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <code>true</code> if it is possible to divide this array into <code>k</code> non-empty subsets whose sums are all equal.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,3,2,3,5,2,1], k = 4
<strong>Output:</strong> true
<strong>Explanation:</strong> It is possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4], k = 3
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= nums.length &lt;= 16</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li>The frequency of each element is in the range <code>[1, 4]</code>.</li>
</ul>

using LL = long long;
class Solution {
public:
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int sum = reduce(nums.begin(), nums.end(), 0);
        
        if (sum % k != 0) return false;
        int target = sum / k;

        unordered_map<LL, bool> memo;

        function<bool(int, int, int)> dp = [&] (int bitmask, int subSum, int k) {
            if (k == 1) {
                return true;
            }

            LL hash = (LL)subSum << 32;
            hash |= bitmask;
            hash <<= 16;
            hash |= k;

            if (subSum == target) {
                return memo[hash] |= dp(bitmask, 0, k - 1);
            }

            if (memo.contains(hash)) return memo[hash];
            memo[hash] = false;

            for (int i = 0 ; i < nums.size() ; ++i ) {
                if (((bitmask & (1 << i)) >> i) == 0 && subSum + nums[i] <= target) {
                    bitmask |= (1 << i);
                    memo[hash] |= dp(bitmask, subSum + nums[i], k);
                    bitmask ^= (1 << i);
                }
            }

            return memo[hash];
        };

        return dp(0, 0, k);
    }
};
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