<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:
    def canPartitionKSubsets(self, nums, k):
        total_sum = sum(nums)
        if total_sum % k != 0:
            return False
        
        target = total_sum // k
        n = len(nums)
        nums.sort(reverse=True)  # Sorting helps in quicker pruning of invalid states

        # DP array with all states initially undefined (using a dictionary)
        dp = {}

        # Helper function for the DP solution
        def search(used, current_sum):
            # If all subsets are formed
            if used == (1 << n) - 1:
                return True
            
            # If the state has been computed before, return it
            if (used, current_sum) in dp:
                return dp[(used, current_sum)]

            # Try to place each element in the current subset
            for i in range(n):
                if not (used & (1 << i)) and current_sum + nums[i] <= target:
                    # Mark the current element as used
                    if search(used | (1 << i), (current_sum + nums[i]) % target):
                        dp[(used, current_sum)] = True
                        return True
            
            # If no valid subset formation is possible, store and return False
            dp[(used, current_sum)] = False
            return False

        # Start with an empty set of used elements and initial subset sum of 0
        return search(0, 0)

# Example usage
solution = Solution()

nums1 = [4, 3, 2, 3, 5, 2, 1]
k1 = 4
print(solution.canPartitionKSubsets(nums1, k1))  # Output: True

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