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The problem at hand is about calculating the Fibonacci numbers, which is a classic problem. Given an integer n, we need to compute the n-th Fibonacci number. The Fibonacci sequence is defined by the following recurrence relation:

• F(0) = 0
• F(1) = 1
• F(n) = F(n - 1) + F(n - 2) for n > 1

Clarifying Questions

1. Q: Should we consider very large values of n? A: The constraints typically ensure that n is a manageable size; for LeetCode, it’s usually within the range [0, 30].
2. Q: Are negative values of n possible in the input? A: No, n is guaranteed to be a non-negative integer.

Strategy

1. Iterative Approach: Use a bottom-up approach to compute the Fibonacci number iteratively. This helps in reducing the space complexity significantly.
2. Dynamic Programming (DP): Store the previously computed Fibonacci numbers to avoid redundant calculations.

Code

Here is an implementation using the iterative approach:

def fib(n: int) -> int:
    if n == 0:
        return 0
    elif n == 1:
        return 1
    
    # Initial base values for the sequence
    a, b = 0, 1
    
    # Iteratively compute the nth Fibonacci number
    for _ in range(2, n + 1):
        a, b = b, a + b
    
    return b

# Example Usage
print(fib(0))  # Output: 0
print(fib(1))  # Output: 1
print(fib(2))  # Output: 1
print(fib(3))  # Output: 2
print(fib(10))  # Output: 55

Time Complexity

• Time Complexity: (O(n))
  • We iterate from 2 to n, so the time complexity is linear with respect to n.
• Space Complexity: (O(1))
  • We only use a constant amount of extra space (variables a and b) regardless of the input size.

This approach ensures that we compute the Fibonacci number efficiently with minimal space usage.

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