# Solving Leetcode Interviews in Seconds with AI: Find the Minimum Number of Fibonacci Numbers Whose Sum Is K


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1414" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the solution, explained, and implemented in Python:

*   **High-Level Approach:**
    *   Generate Fibonacci numbers up to `k`.
    *   Greedily subtract the largest Fibonacci number less than or equal to the remaining `k` until `k` becomes 0.
    *   Count the number of Fibonacci numbers used.

*   **Complexity:**
    *   Runtime: O(log k). The Fibonacci sequence grows exponentially, so the number of Fibonacci numbers generated and the number of iterations in the greedy subtraction are logarithmic with respect to k.
    *   Storage: O(log k) to store the generated Fibonacci numbers.

	
	# Code
	```python
	def findMinFibonacciNumbers(k: int) -> int:
    """
    Finds the minimum number of Fibonacci numbers whose sum is equal to k.

    Args:
        k: The target integer.

    Returns:
        The minimum number of Fibonacci numbers.
    """

    fibonacci_numbers = [1, 1]
    while fibonacci_numbers[-1] < k:
        next_fib = fibonacci_numbers[-1] + fibonacci_numbers[-2]
        fibonacci_numbers.append(next_fib)

    count = 0
    i = len(fibonacci_numbers) - 1
    while k > 0:
        if fibonacci_numbers[i] <= k:
            k -= fibonacci_numbers[i]
            count += 1
        i -= 1

    return count
	```