<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 {
private:
    bool solve(int i, int mask, int &n, int &sum, vector<int> &nums, int &k, vector<vector<int>> &dp)
    {
        if(i == k)
        {
            if(mask == ((1 << n) - 1))
                return true;
            return false;
        }
        if(dp[i][mask] != -1)
            return dp[i][mask];
        int curr_sum = 0;
        for(int j = 0; j < n; j++)
            if(mask & (1 << j))
                curr_sum += nums[j];
        curr_sum -= i * sum;
        bool ans = false;
        for(int j = 0; j < n; j++)
        {
            if(mask & (1 << j))
                continue;
            if(nums[j] + curr_sum < sum)
                ans = ans or solve(i, mask | (1 << j), n, sum, nums, k, dp);
            else if(nums[j] + curr_sum == sum)
                ans = ans or solve(i + 1, mask | (1 << j), n, sum, nums, k, dp);
        }
        return dp[i][mask] = ans;
    }
public:
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);

        int sum = 0;
        for(int num: nums)
            sum += num;
        if(sum % k or *max_element(nums.begin(), nums.end()) > sum)
            return false;
        
        sum /= k;
        int n = nums.size();
        vector<vector<int>> dp(k, vector<int> (1 << n, -1));
        return solve(0, 0, n, sum, nums, k, dp);
    }
};
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