You are given the following problem:
The Fibonacci numbers, commonly denoted F(n)
form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0
and 1
. That is,
F(0) = 0, F(1) = 1
F(n) = F(n - 1) + F(n - 2), for n > 1.
Given n
, calculate F(n)
. Let's look at some examples:
n = 2
, the expected output is 1
because F(2) = F(1) + F(0) = 1 + 0 = 1
.n = 3
, the expected output is 2
because F(3) = F(2) + F(1) = 1 + 1 = 2
.n = 4
, the expected output is 3
because F(4) = F(3) + F(2) = 2 + 1 = 3
.Write a function that takes an integer n
as input (where 0 <= n <= 30
) and returns the nth Fibonacci number.
How would you implement this function, and what are the time and space complexities of your solution? Also, consider any edge cases that might be encountered.
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The Fibonacci sequence is a series of numbers where each number is the sum of the two numbers before it. The brute force approach directly applies the definition to calculate the sequence, without trying to optimize the calculation process.
Here's how the algorithm would work step-by-step:
def fibonacci_brute_force(number_to_find):
# Base cases: first and second Fibonacci numbers are 1
if number_to_find <= 2:
return 1
# Recursively calculate the Fibonacci number.
previous_fibonacci_number = fibonacci_brute_force(number_to_find - 1)
penultimate_fibonacci_number = fibonacci_brute_force(number_to_find - 2)
# The fibonacci number is the sum of the prior two.
return previous_fibonacci_number + penultimate_fibonacci_number
To find a specific Fibonacci number quickly, we avoid recalculating values we've already found. Instead, we store previously computed Fibonacci numbers and reuse them as needed, building up to the number we're looking for.
Here's how the algorithm would work step-by-step:
def fibonacci_number(n):
if n <= 1:
return n
previous_fib_number = 0
current_fib_number = 1
# We iterate to build up to the nth Fibonacci number
for _ in range(2, n + 1):
next_fib_number = previous_fib_number + current_fib_number
# We are storing values, shifting the 'window' forward
previous_fib_number = current_fib_number
# This updates current to be the sum just calculated.
current_fib_number = next_fib_number
return current_fib_number
Case | How to Handle |
---|---|
Input n is zero | Return 0, as F(0) = 0 is defined as the base case. |
Input n is one | Return 1, as F(1) = 1 is defined as the second base case. |
Input n is a large positive integer | Use dynamic programming or memoization to avoid stack overflow and potential integer overflow when using recursion. |
Negative input n | Return an error or define Fibonacci for negative indices using F(-n) = (-1)^(n+1)F(n). |
Integer overflow for large n using iterative solution | Use a larger data type (e.g., long long in C++, BigInteger in Java/Python) or return an error if overflow is unavoidable. |
Recursive solution with large n exceeds maximum recursion depth | Convert to an iterative (dynamic programming) approach instead of recursion. |
Floating-point errors in Binet's formula | Avoid using Binet's formula due to floating-point imprecision; favor dynamic programming or iterative solutions. |
Memoization array size for large n exceeds memory limits | Consider a space-optimized dynamic programming approach storing only the last two Fibonacci numbers, rather than an entire array. |