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The Question
Coding

Climbing Stairs

You are climbing a staircase that takes n steps to reach the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? Constraints: - 1 <= n <= 45 - The solution should handle the result within a 32-bit signed integer range.
C++
DP
Fibonacci Sequence
Questions & Insights

Clarifying Questions

What is the maximum value of $n$? (Assuming n \le 45 based on standard constraints, as n=46 exceeds the 32-bit signed integer limit).
Are the number of steps always 1 or 2? (Assuming yes, as per the standard problem definition).
What is the base case for $n=0$ or $n=1$? (Assuming n \ge 1; if n=0, usually there is 1 way—doing nothing—but standard constraints start at 1).
Is space complexity a concern? (The problem can be solved in O(n) or O(1) space; I will provide the O(1) optimization).
Assumptions:
n is a positive integer.
Result fits within a 32-bit signed integer for n \le 45.
Performance target is O(n) time.

Thinking Process

Identify the Overlapping Subproblems: To reach step i, you must have come from either step i-1 (by taking 1 step) or step i-2 (by taking 2 steps). Thus, the total ways to reach is the sum of ways to reach i-1 and i-2.
Recognize the Pattern: This recurrence relation dp[i] = dp[i-1] + dp[i-2] is identical to the Fibonacci sequence, where dp[1] = 1 and dp[2] = 2.
Optimize Space: Instead of maintaining an entire array of size n, we only need the last two calculated values to compute the current value. This reduces space complexity from O(n) to O(1).
Handle Edge Cases: Ensure n=1 and n=2 are handled correctly as they form the foundation of the sequence.
Implementation Breakdown

Problem Set

Functional Requirement: Given an integer n, return the number of distinct ways to climb to the top.
Constraints:
1 \le n \le 45.
Time complexity should be O(n).
Space complexity should be O(1).

Approach

Algorithm: Dynamic Programming (Bottom-Up Iterative)
Data Structure: Two variables (Scalar state reduction)
Complexity:
Time: O(n)
Space: O(1)

Implementation

Wrap Up

Advanced Topics

Matrix Exponentiation: This problem can be solved in O(\log n) time using matrix exponentiation of the Fibonacci transformation matrix \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n. This is useful if n is extremely large (though we would need to return the result modulo M as it would exceed 64-bit limits).
Closed-form Solution (Binet's Formula): The n-th Fibonacci number can be calculated in O(1) using the golden ratio \phi, but floating-point precision issues make it unreliable for large n in competitive programming.
Generic Step Sizes: If the allowed steps were a set S = \{s_1, s_2, ..., s_k\}, the complexity would become O(n \cdot |S|). This is a variation of the "Change Making Problem" or "Coin Change".