<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 {
    int k, target;
    vector<int> memo; 
    
    bool partition(vector<int>& nums, int mask, int sum, int targetReached) {
        if (sum > target) return false;
        if (targetReached == k) return true;  
        if (sum == target)  
            return partition(nums, mask, 0, targetReached + 1);
        
        if (memo[mask] != -1) return memo[mask];
        
        for (int i = 0; i < nums.size(); i++) {
            if (!(mask & (1 << i))) { 
                if (partition(nums, mask | (1 << i), sum + nums[i], targetReached)) {
                    return memo[mask] = true;
                }
            }
        }
        return memo[mask] = false;
    }
    
public:
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        this->k = k;
        int totalSum = accumulate(nums.begin(), nums.end(), 0);
        if (totalSum % k != 0) return false; 
        
        sort(nums.begin(), nums.end(), greater<int>());
        target = totalSum / k;
        if (nums[0] > target) return false;
        
        int n = nums.size();
        memo = vector<int>(1 << n, -1);
        return partition(nums, 0, 0, 0);
    }
};

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