Find minimum subset that sum to at least K | Yaozong's Blog


Find minimum subset that sum to at least K | Yaozong's Blog
Given a list of n integers, write a function that outputs the minimum subset of numbers that sum to at least K.
Follow up: can you beat O(nlogn)?
Solution 1: sort and linear search  O(nlogn)


vector
<int> minimum_subset_sum_atleast_K(vector<int>& vec, int K)
{
  if(vec.size() == 0)
      return {};
  
  sort(vec.begin(), vec.end());
  int n = vec.size();
  
  vector<int> res(1, vec.back());
  int cumsum = vec.back();
  for(int i = n - 2; i >= 0; --i)
  {
    if(vec[i] < 0 || cumsum >= K) break;
    cumsum += vec[i];
    res.push_back(vec[i]);
  }
  
  if(cumsum >= K)
      return res;
  else
      return {};
}
Here we introduce a selection based algorithm which on average costs O(n) but costs O(n^2) in the worst case. The approach is similar to kth order-statistics.

  • Randomly choose a pivot from the current array and partition the array into two parts p1 and p2p1 contains integers which are smaller than the pivot; p2 contains integers which are larger than or equal to the pivot.
  • Check if pivot + sum(p2) == K, if so, we already find the answer; if not, then if pivot + sum(p2) > K but sum(p2) <= K, we also find the answer; if  sum(p2) > K, we need to do further partition, but we just need to further partition p2;  if pivot + sum(p2) < K,  we need to do further partition, but we need to partition p1 instead and the target value K is updated as K - pivot - sum(p2).
void find_min_set(int *a, int start, int end, int target)
{

    /* generate secret number: */
    int s, e, k;

    s = start;
    e = end;
    k = target;

    int sum;

    while(1)
    {
       srand ( time(NULL) );
       int idx = ((int)rand()) % (e-s + 1) + s;
       
       /* partition and sum */
       int tmp = a[idx];
       a[idx] = a[e];
       a[e] = tmp;

       sum = a[e];
       int pos = s;
       for(int i = s; i < e; i++)
       {
          if(a[i] < a[e])
          {
             if(pos != i)
             {
                tmp = a[pos];
                a[pos] = a[i];
                a[i] = tmp;
             }
            pos++;
          }
          else sum+= a[i];

       }

      tmp = a[pos];
      a[pos] = a[e];
      a[e] = tmp;

      if(sum == k)
      {
        cout << "found " << endl;
        return;
      }
      else if(sum > k)
      {
         if(sum - a[pos] < k)
         {
            cout << "found " << endl;
            return;
         }
         s = pos + 1;
         sum = 0;

      }
      else
      {
        k = k - sum;
        sum = 0;
        e = pos - 1;
      }
    }

}
vector<int> quick_select(vector<int>& vec, int i, int j, int K)
{
  if(i > j)
    return {};
  if(i == j)
    return {vec[i]};
  
  int len = j - i + 1;
  int idx = rand() % len + i;
  int pivot = vec[idx];
  
  swap(vec[idx], vec[j]);
  
  int p = i, q = j;
  int left_sum = vec[idx], right_sum = 0;
  
  while(p < q) {
    
    if(vec[p] <= pivot) {
        left_sum += vec[p];
        ++ p;
    } else {
      right_sum += vec[p];
      swap(vec[p], vec[--q]);
    }
      
  }
  
  swap(vec[q], vec[j]);
  
  vector<int> res;
  if(right_sum == K) {
    if(q + 1 <= j)
      res.insert(res.end(), vec.begin()+q+1, vec.begin()+j+1);
    return res;
  }
  else if(right_sum > K) {
    return quick_select(vec, q+1, j, K);
  }
  else {
    if(q + 1 <= j)
      res.insert(res.end(), vec.begin()+q+1, vec.begin()+j+1);
    K = K - right_sum;
    vector<int> res2 = quick_select(vec, i, q, K);
    res.insert(res.end(), res2.begin(), res2.end());
    return res;
  }
  
}
vector<int> minimum_subset_sum_atleast_K(vector<int>& vec, int K)
{
  if(vec.size() == 0)
      return {};
  vector<int> res = quick_select(vec, 0, vec.size() - 1, K);
  int sum = accumulate(res.begin(), res.end(), 0);
  if(sum >= K)
    return res;
  else
    return {};
}

https://www.quora.com/Find-the-maximum-sum-of-k-length-subset-of-a-given-set-such-that-sum-is-strictly-less-than-M
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