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To solve the problem of finding the number of distinct ways to reach the nth step (taking 1 or 2 steps at a time), we can leverage a Fibonacci-like recurrence relation since each step depends on the previous two steps.

Approach

The key insight is:

• To reach the nth step, the last move could either be a 1-step from the (n-1)th step or a 2-step from the (n-2)th step.
• Thus, the number of ways to reach step n is the sum of the ways to reach step n-1 and step n-2.

This leads to the recurrence relation:
f(n) = f(n-1) + f(n-2)

Base cases:

• f(1) = 1 (only one way: take a single step)
• f(2) = 2 (two ways: 1+1 or directly 2 steps)

The sequence generated is: 1,2,3,5,8,... which is the Fibonacci sequence shifted by one position (f(n) = Fib(n+1), where Fib(1)=1, Fib(2)=1).

Solution Code (Iterative)

This approach uses O(n) time and O(1) space (optimal for most cases):

def climb_stairs(n):
    if n == 1:
        return 1
    elif n == 2:
        return 2
    a, b = 1, 2  # a = f(n-2), b = f(n-1)
    for _ in range(3, n+1):
        c = a + b
        a = b
        b = c
    return b

Explanation

1. Base Cases: Directly return results for small values of n (1 or 2).
2. Iterative Calculation: For n >=3, we maintain two variables (a and b) to store the number of ways to reach the previous two steps. We update these variables iteratively until we reach the nth step.

Example:
For n=5:

• Start with a=1 (f(1)), b=2 (f(2))
• n=3: c=1+2=3 → a=2, b=3
• n=4: c=2+3=5 → a=3, b=5
• n=5: c=3+5=8 → return b=8

Fast Doubling Method (O(log n) Time)

For very large n (e.g., 1e18), use the fast doubling technique to compute Fibonacci numbers in logarithmic time:

def fib(n):
    if n == 0:
        return (0, 1)
    a, b = fib(n >> 1)  # n//2
    c = a * (2*b - a)
    d = a*a + b*b
    if n & 1:  # n is odd
        return (d, c + d)
    else:
        return (c, d)

def climb_stairs_fast(n):
    return fib(n)[1]  # since f(n) = Fib(n+1) → fib(n+1)[0]? Wait let's confirm:
    # Wait, let's adjust: if fib(n) returns (Fib(n), Fib(n+1)), then Fib(n+1) is the second element of fib(n). 
    # For f(1)=1=Fib(2) → fib(1) returns (Fib(1), Fib(2)) → (1, 2)? Let's see:
    # Let's recheck: the fast doubling function here: let's take n=1:
    # n=1 is odd, n>>1=0 → fib(0) returns (0, 1). Then c=0*(2*1 - 0)=0; d=02+12=1. Since n is odd, return (d, c+d) → (1,0+1=1). So Fib(1)=1, Fib(2)=1. Then f(n)= Fib(n+1). So f(1)= Fib(2)=1 → fib(1)[1] → yes (fib(1) returns (1, 1)). f(2)= Fib(3)=2 → fib(2) returns what? n=2 even: n>>1=1 → fib(1) returns (1, 1). c= 1*(2*1-1)=1*1=1; d= 1+1=2. So fib(2) returns (1,2). So Fib(3)=2 → which is fib(2)[1]. So yes: f(n)= fib(n)[1]. Because f(n)= Fib(n+1) → fib(n) gives (Fib(n), Fib(n+1)). So yes, climb_stairs_fast(n) = fib(n)[1].

Key Takeaways

• The problem is a classic example of dynamic programming (overlapping subproblems and optimal substructure).
• The iterative approach is best for most practical scenarios (simple and efficient).
• For extremely large n, use the fast doubling method to reduce time complexity to O(log n).

Choose the method based on the constraints of your problem!

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