<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 solve(int index, vector<int>& nums, int n, int k, int target, vector<int>& subsetSum, vector<vector<int>>& dp) {
        if (index == n)
            return true; // Successfully placed all elements into subsets
        
        for (int i = 0; i < k; ++i) {
            if (subsetSum[i] + nums[index] <= target) {
                subsetSum[i] += nums[index];
                if (solve(index + 1, nums, n, k, target, subsetSum, dp))
                    return true;
                subsetSum[i] -= nums[index];
            }
            // If a subset sum is still zero, break to avoid redundant permutations
            if (subsetSum[i] == 0) break;
        }

        return false;
    }

    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int total = accumulate(nums.begin(), nums.end(), 0);
        
        if (total % k != 0) 
            return false;

        int target = total / k;
        vector<int> subsetSum(k, 0);
        vector<vector<int>> dp(nums.size(), vector<int>(total + 1, -1));
        
        sort(nums.rbegin(), nums.rend()); // Sort in descending order for optimization
        
        return solve(0, nums, nums.size(), k, target, subsetSum, dp);
    }
};

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