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
func canPartitionKSubsets(nums []int, k int) bool {
	s := 0
	for _, v := range nums {
		s += v
	}
	if s%k != 0 {
		return false
	}
	s /= k
	cur := make([]int, k)
	n := len(nums)

	var dfs func(int) bool
	dfs = func(i int) bool {
		if i == n {
			return true
		}
		for j := 0; j < k; j++ {
			if j > 0 && 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.Sort(sort.Reverse(sort.IntSlice(nums)))
	return dfs(0)
}