<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:
    // https://leetcode.com/problems/partition-to-k-equal-sum-subsets/solutions/480707/c-dp-bit-manipulation-in-20-lines

    vector<int> dp;

    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int n = nums.size(), sum = 0;
        dp.resize((1<<16), -1);
        dp[0] = 0;
        for (int i = 0; i < n; i++)
            sum += nums[i];
        if (sum % k != 0)
            return false;

        int tar = sum/k;
        // Please note n < 16
        for (int mask = 0; mask < (1<<n); mask++) { // 2^n (covers all combinations on n numbers)
            if (dp[mask] == -1) continue;  // if current state is illegal, simply ignore it
             for (int i = 0; i < n; i++) {
                if (!(mask&(1<<i)) && dp[mask]+nums[i] <= tar) {  // if nums[i] is unchosen && choose nums[i] would not cross the target
                    dp[mask|(1<<i)] = (dp[mask]+nums[i]) % tar;
                }
             }
        }
        return dp[(1<<n)-1] == 0;
    }
};
58934