<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 n = nums.size();
        int total_sum = 0;
        for(int i = 0; i < n; i++) {
            total_sum += nums[i];
        }
        
        if (total_sum % k != 0) return false;
        
        int target = total_sum / k;
        vector<long long> dp((1 << 16) + 2, -1);
        dp[0] = 0;
        
        for (int mask = 0; mask < (1<<n); mask++) {
            if (dp[mask] == -1) continue;
            
            for (int i = 0; i < n; i++) {
                if (!(mask & (1 << i)) && dp[mask] + nums[i] <= target) {
                    dp[mask | (1 << i)] = (dp[mask] + nums[i]) % target;
                }
            }
        }
        
        return (dp[(1 << n) - 1] == 0);
    }
};
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