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
function canPartitionKSubsets(nums: number[], k: number): boolean {
    const dfs = (i: number): boolean => {
        if (i === nums.length) {
            return true;
        }
        for (let 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;
    };

    let s = nums.reduce((a, b) => a + b, 0);
    const mod = s % k;
    if (mod !== 0) {
        return false;
    }
    s = Math.floor(s / k);
    const cur = Array(k).fill(0);
    nums.sort((a, b) => b - a);
    return dfs(0);
}
