Formatted question description: https://leetcode.ca/all/1414.html

1414. Find the Minimum Number of Fibonacci Numbers Whose Sum Is K (Medium)

Given the number k, return the minimum number of Fibonacci numbers whose sum is equal to k, whether a Fibonacci number could be used multiple times.

The Fibonacci numbers are defined as:

  • F1 = 1
  • F2 = 1
  • Fn = Fn-1 + Fn-2 , for n > 2.

It is guaranteed that for the given constraints we can always find such fibonacci numbers that sum k.

 

Example 1:

Input: k = 7
Output: 2 
Explanation: The Fibonacci numbers are: 1, 1, 2, 3, 5, 8, 13, ... 
For k = 7 we can use 2 + 5 = 7.

Example 2:

Input: k = 10
Output: 2 
Explanation: For k = 10 we can use 2 + 8 = 10.

Example 3:

Input: k = 19
Output: 3 
Explanation: For k = 19 we can use 1 + 5 + 13 = 19.

 

Constraints:

  • 1 <= k <= 10^9

Related Topics:
Array, Greedy

Solution 1. Greedy

We greedily find the maximum fibonacci number that is smaller or equal to k.

Assume it’s b, then f(k) = f(k-b) + 1. f(0) = 0.

// OJ: https://leetcode.com/problems/find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k/

// Time: O(F(k)) where F(k) is the steps to compute the fibonacci number greater than k
// Space: O(1)
class Solution {
public:
    int findMinFibonacciNumbers(int k) {
        if (k == 0) return 0;
        int a = 1, b = 1;
        while (a + b <= k) {
            int c = a + b;
            a = b;
            b = c;
        }
        return findMinFibonacciNumbers(k - b) + 1;
    }
};

Java

  • class Solution {
        public int findMinFibonacciNumbers(int k) {
            TreeSet<Integer> set = new TreeSet<Integer>();
            set.add(0);
            set.add(1);
            int prev2 = 1, prev1 = 1;
            int num = 2;
            while (num <= k) {
                set.add(num);
                prev2 = prev1;
                prev1 = num;
                num = prev2 + prev1;
            }
            int count = 0;
            while (k > 0) {
                int fibo = set.floor(k);
                k -= fibo;
                count++;
            }
            return count;
        }
    }
    
  • // OJ: https://leetcode.com/problems/find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k/
    // Time: O(F(k)) where F(k) is the steps to compute the fibonacci number greater than k
    // Space: O(1)
    class Solution {
    public:
        int findMinFibonacciNumbers(int k) {
            if (k == 0) return 0;
            int a = 1, b = 1;
            while (a + b <= k) {
                int c = a + b;
                a = b;
                b = c;
            }
            return findMinFibonacciNumbers(k - b) + 1;
        }
    };
    
  • class Solution:
        def findMinFibonacciNumbers(self, k: int) -> int:
            fib = [1, 1]
            while fib[-1] < k:
                fib.append(fib[-1] + fib[-2])
            res = 0
            while k != 0:
                for i in range(len(fib) - 1, -1, -1):
                    if fib[i] <= k:
                        k -= fib[i]
                        res += 1
                        break
            return res
    

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