<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: List[int], k: int) -> bool:
        # this is probably exponential searching the space for the divisons
        # possible in 4 steps each taking the result

        total = sum(nums)
        if total % k != 0:
            return False # floats can't be divided into subsets
        target = total // k
        nums = sorted(nums,reverse=True)
        @lru_cache(None)
        def r(b,current,step):
            # we forgo the need for the last division since if previous ones correct this is given
            if step == k-1:
                return True
            if b == 0: # we haven't partitioned and no more selections
                return False
            
            result = False
            for i in range(len(nums)):
                if (b & (1<<i)):
                    cstep, csum = step, current + nums[i]
                    if current + nums[i] == target:
                        cstep += 1 # if we reach a target we just step next
                        csum = 0
                    elif csum > target:
                        continue # early termination
                    result = result or r(b & ~(1<<i), csum, cstep)
            
            # if we overshoot the sum, it'll fall through and we never complete the steps
            return result

        bits = ~((~0)<<(len(nums)+1))  
        return r(bits, 0, 0)
        
        

            
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