<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:
    unordered_map<int, unordered_map<int, bool>> dp;  // dp[mask][curSum]

    bool canPartition(int mask, int curSum, int count, int k, vector<int>& nums, int target) {
        if (count == k - 1) {  // If we've formed k-1 subsets, the last one must be valid
            return true;
        }
        
        if (curSum == target) {  // If current subset sum reaches target, start a new subset
            return canPartition(mask, 0, count + 1, k, nums, target);
        }

        if (dp.find(mask) != dp.end() && dp[mask].find(curSum) != dp[mask].end()) {
            return dp[mask][curSum];
        }

        for (int i = 0; i < nums.size(); i++) {
            if ((mask & (1 << i)) == 0 && curSum + nums[i] <= target) {  // Try to include nums[i] in the current subset
                if (canPartition(mask | (1 << i), curSum + nums[i], count, k, nums, target)) {
                    return dp[mask][curSum] = true;
                }
            }
        }

        return dp[mask][curSum] = false;  // Memoize result
    }

    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int totalSum = accumulate(nums.begin(), nums.end(), 0);
        if (totalSum % k != 0) {
            return false;  // If the total sum isn't divisible by k, partitioning is impossible
        }
        
        int target = totalSum / k;  // Each subset must sum to this value
        int n = nums.size();

        dp.clear();  // Clear the memoization map
        sort(nums.begin(), nums.end(), greater<int>());  // Sort in descending order for optimization
        if (nums[0] > target) return false;  // If the largest element is greater than the target, it's impossible
        
        return canPartition(0, 0, 0, k, nums, target);  // Start the recursion with an empty subset
    }
};

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