<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 {
    unordered_map<int, bool> memo;

    bool backtrack(vector<int>& nums, int k, int bucketSum, int start, int used, int target) {
        if(k==0) return true;
        if(bucketSum > target) return false;

        if(memo.count(used)) {
            return memo[used];
        }

        if(bucketSum == target) {
            return memo[used] = backtrack(nums, k-1, 0, 0, used, target);
        }
        if(start == nums.size()) return false;

        bool isPossible = false;
        if(((used >> start) & 1) == 0) {
            used |= (1<<start);
            isPossible = backtrack(nums, k, bucketSum+nums[start], start+1, used, target);
            if(isPossible) return memo[used] = true;
            used ^= (1<<start);
        }
        isPossible |= backtrack(nums, k, bucketSum, start+1, used, target);

        // for(int i=start; i<nums.size(); ++i) {
        //     if(((used >> i) & 1) == 1) continue;

        //     used |= (1<<i);
        //     bucketSum += nums[i];
        //     if(backtrack(nums, k, bucketSum, i+1, used, target)) {
        //         return memo[used] = true;
        //     }
        //     bucketSum -= nums[i];
        //     used ^= (1<<i);
        // }

        return memo[used] = isPossible;
    }
public:
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        if(k > nums.size()) return false;
        int sum = accumulate(nums.begin(), nums.end(), 0);
        if(sum%k != 0) return false;

        int target = sum/k;
        int used = 0;
        return backtrack(nums, k, 0, 0, used, target);
    }
};

/*

n nums, k buckets

sum of each bucket = 20/4=5

1. from nums prespective: O(k*k...k) = O(k^n)
2. from bucket prespective: O((2*2*..2)*k) = O(2^n*k)

2nd approach is efficient

2^16

*/
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