To solve this problem, we need to compute the nth Fibonacci number based on the given sequence pattern. The sequence follows the Fibonacci series where each number is the sum of the two preceding ones, starting from 1 and 1.

Approach

The Fibonacci sequence for this problem is defined as:

• Fib(1) = 1
• Fib(2) = 1
• Fib(n) = Fib(n-1) + Fib(n-2) for n > 2

We use an iterative approach to compute the Fibonacci number efficiently, avoiding the stack overflow issues associated with recursive methods. This approach runs in O(n) time complexity, which is optimal for this problem.

Solution Code

def compute_fibonacci(n):
    if n <= 1:
        return n
    a, b = 1, 1
    for _ in range(2, n):
        a, b = b, a + b
    return b

# Example usage:
# print(compute_fibonacci(2))  # Output: 1
# print(compute_fibonacci(3))  # Output: 2
# print(compute_fibonacci(4))  # Output: 3
# print(compute_fibonacci(5))  # Output:5
# print(compute_fibonacci(6))  # Output:8

Explanation

1. Base Cases: For n=1 or n=2, the result is directly 1.
2. Iterative Calculation: We initialize two variables a and b to 1 (representing Fib(1) and Fib(2)). For each number from 3 to n, we update a and b such that a takes the value of b and b takes the sum of the previous a and b. This way, we build up the Fibonacci sequence up to the nth term.

This method efficiently computes the desired Fibonacci number with minimal computational overhead, making it suitable for most practical values of n. If very large values of n (like 1e18) are required, we can switch to faster methods like matrix exponentiation, but the iterative approach works well for typical problem constraints.

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