<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 {
// private:
//     bool backtrack(vector<int>& nums, int index, int count, int currSum, int k, int subSum, std::string taken, std::unordered_map<std::string, bool> memo){
//         int numsSize = nums.size();
//         if(count == k - 1){
//             return true;
//         }
//         if(currSum > subSum){
//             return false;
//         }
//         if(memo.find(taken) != memo.end()){
//             return memo[taken];
//         }
//         if(currSum == subSum){
//             return backtrack(nums, 0, count + 1, 0 , k, subSum, taken, memo);
//         }
//         for(int j = index; j < numsSize; j++){
//             if(taken[j] == '0'){
//                 taken[j] = '1';
//                 if(backtrack(nums, j + 1, count, currSum + nums[j], k, subSum, taken, memo)){
//                     return true;
//                 }
//                 taken[j] = '0';
//             }

//         }
//         return memo[taken] = false;
//     }
// public:
//     bool canPartitionKSubsets(vector<int>& nums, int k) {
//         int sum = 0;
//         int numsSize = nums.size();

//         for(int i = 0; i < numsSize; i++){
//             sum += nums[i];
//         }
//         if(sum % k){
//             return false;
//         }
//         int subSum = sum / k;

//         // sort the nums in decreasing order
//         std::sort(nums.begin(), nums.end(), std::greater<int>());

//         std::string taken(numsSize, '0');
//         std::unordered_map<std::string, bool> memo;
//         return backtrack(nums, 0, 0 ,0, k, subSum, taken, memo);
        
//     }
// };

class Solution {
public:
    bool backtrack(vector<int>& arr, int index, int count, int currSum, int k, 
                   int targetSum, string& taken, unordered_map<string, bool>& memo) {

        int n = arr.size();
      
        // We made k - 1 subsets with target sum and last subset will also have target sum.
        if (count == k - 1) { 
            return true;
        }
        
        // No need to proceed further.
        if (currSum > targetSum) { 
            return false;
        }
        
        // If we have already computed the current combination.
        if (memo.find(taken) != memo.end()) {
            return memo[taken];
        }
      
        // When curr sum reaches target then one subset is made.
        // Increment count and reset current sum.
        if (currSum == targetSum) {
            return memo[taken] = backtrack(arr, 0, count + 1, 0, k, targetSum, taken, memo);
        }
        
        // Try not picked elements to make some combinations.
        for (int j = index; j < n; ++j) {
            if (taken[j] == '0') {
                // Include this element in current subset.
                taken[j] = '1';
                
                // If using current jth element in this subset leads to make all valid subsets.
                if (backtrack(arr, j + 1, count, currSum + arr[j], k, targetSum, taken, memo)) {
                    return true;
                }
                
                // Backtrack step.
                taken[j] = '0';
            }
        } 
      
        // We were not able to make a valid combination after picking each element from the array,
        // hence we can't make k subsets.
        return memo[taken] = false;
    }
  
    bool canPartitionKSubsets(vector<int>& arr, int k) {
        int totalArraySum = 0;
        int n = arr.size();

        for (int i = 0; i < n; ++i) {
             totalArraySum += arr[i];
        }
      
        // If total sum not divisible by k, we can't make subsets.
        if (totalArraySum % k != 0) { 
            return false;
        }
      
        // Sort in decreasing order.
        sort(arr.begin(), arr.end(), greater<int>());
        
        int targetSum = totalArraySum / k;
        
        // String to represent taken array.
        string taken(n, '0');
        
        // Memoize the ans using taken element's string as key.
        unordered_map<string, bool> memo;
      
        return backtrack(arr, 0, 0, 0, k, targetSum, taken, memo);
    }
};
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