Partition to K Equal Sum Subsets
<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>
Bit Manipulation; Memoization; Array; Dynamic Programming; Backtracking; Bitmask
class Solution {
public:
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int s = accumulate(nums.begin(), nums.end(), 0);
        if (s % k) {
            return false;
        }
        s /= k;
        int n = nums.size();
        vector<int> cur(k);
        function<bool(int)> dfs = [&](int i) {
            if (i == n) {
                return true;
            }
            for (int j = 0; j < k; ++j) {
                if (j && cur[j] == cur[j - 1]) {
                    continue;
                }
                cur[j] += nums[i];
                if (cur[j] <= s && dfs(i + 1)) {
                    return true;
                }
                cur[j] -= nums[i];
            }
            return false;
        };
        sort(nums.begin(), nums.end(), greater<int>());
        return dfs(0);
    }
};
