README.md

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
• F_n = F_n-1 + F_n-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/
// Author: github.com/lzl124631x
// 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;
    }
};