<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>

class Solution {
public:
    bool canPartitionKSubsets(vector<int>& arr, int k) {
        int s(0), n(arr.size());
        for (int i(0); i < n; ++i) s += arr[i];
        if (s%k) return 0;
        sort(arr.rbegin(), arr.rend());
        int des(s/k);
        string taken(n, '0');
        map<string, bool> dp;
        function<bool(int, int, int, int)> backtrack = [&](int index, int count, int currSum, int k) -> bool {
            if (count == k-1) 
                return 1;
            if (currSum > des) 
                return 0;
            if (dp.find(taken) != dp.end()) 
                return dp[taken];
            if (currSum == des) 
                return dp[taken] = backtrack(0, count+1, 0, k);
            for (int j(index); j < n; ++j) {
                if (taken[j] == '0') {
                    taken[j] = '1';
                    if (backtrack(j+1, count, currSum + arr[j], k)) 
                        return 1;
                    taken[j] = '0';
                }
            }
            return dp[taken] = 0;
        };
        return backtrack(0, 0, 0, k);
    }
};

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