LeetCode in Kotlin

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

Medium

Given an integer k, return the minimum number of Fibonacci numbers whose sum is equal to k. The same Fibonacci number can 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 up to 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 <= 109

Solution

class Solution {
    fun findMinFibonacciNumbers(k: Int): Int {
        val list: MutableList<Int> = mutableListOf(1, 1)
        var prev = 1
        var curr = 1
        while (prev <= k) {
            val n = prev + curr
            prev = curr
            curr = n
            list.add(n)
        }
        var count = 0
        var num = k
        for (i in list.indices.reversed()) {
            if (list[i] <= num) {
                count++
                num = num - list[i]
            }
        }
        return count
    }
}

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