<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>& nums, int k) {
        int sum = 0;
        for (auto a : nums) sum += a;
        if (k <= 0 || sum%k != 0) return false;
        int visited = 0;
        return dfs(nums, visited, 0, k, 0, sum/k);
    }

private:
    unordered_map<string, bool> dp;
    bool dfs(const vector<int>& nums, int& visited, int p, int k, int sum, int target) {
        if (k == 1) return true;
        string t = to_string(visited) + to_string(p) + to_string(k) + to_string(sum);
        if (dp.find(t) != dp.end()) return dp[t];
        if (sum == target) return dfs(nums, visited, 0, k-1, 0, target);
        for (int i = p; i < nums.size(); ++i) {
            if ((visited>>i)&1 || sum + nums[i] > target) continue;
            visited |= (1<<i);
            if (dfs(nums, visited, i+1, k, sum+nums[i], target)) return dp[t] = true;
            visited &= ~(1<<i);
        }
        return dp[t] = false;
    }
};
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