Minimum Number of Fibonacci Numbers Whose Sum is K in C++

Suppose we have a number k, we have to find the minimum number of Fibonacci numbers whose sum is equal to the k, whether a Fibonacci number could be used multiple times.

So, if the input is like k = 7, then the output will be 2, as the Fibonacci numbers are: 1, 1, 2, 3, 5, 8, 13, ... For k = 7 we can use 2 + 5 = 7.

To solve this, we will follow these steps −

• Define an array f

• insert 0 at the end of f

• insert 1 at the end of f

• while last element of f <= k, do −

  • insert (last element of f + second last element of f) into f

• ret := 0

• j := last index of f

• while (j >= 0 and k > 0), do −

  • if f[j] <= k, then −

    • k := k - f[j]

    • (increase ret by 1)

  • Otherwise

    • (decrease j by 1)

• return ret

Example 

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   int findMinFibonacciNumbers(int k) {
      vector<int> f;
      f.push_back(0);
      f.push_back(1);
      while (f.back() <= k) {
         f.push_back(f[f.size() - 1] + f[f.size() - 2]);
      }
      int ret = 0;
      int j = f.size() - 1;
      while (j >= 0 && k > 0) {
         if (f[j] <= k) {
            k -= f[j];
            ret++;
         }
         else
            j--;
      }
      return ret;
   }
};
main(){
   Solution ob;
   cout << (ob.findMinFibonacciNumbers(7));
}

Input

7

Output

2