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Climbing Stairs

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Dynamic Programming

You are climbing a staircase. It 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?

Example 1:

Input: n = 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

Input: n = 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

Constraints:

1 <= n <= 45

Try coding it in LeetCode

Climbing Stairs

Easy

Asked by:

+12

More companies

204 views

Topics:

Dynamic Programming

You are climbing a staircase. It 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?

Example 1:

Input: n = 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

Input: n = 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

Constraints:

1 <= n <= 45

Try coding it in LeetCode

Solution

Clarifying Questions

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:

What are the constraints on the value of 'n', specifically its minimum and maximum possible values?
Is 'n' always a positive integer, or should I consider cases where 'n' might be zero or negative?
Are there any specific performance requirements or time/space complexity targets I should aim for?
Is it possible for 'n' to be so large that the number of distinct ways might exceed standard integer types?
Can 'n' be 1? If so, what is the expected output for n=1?

Brute Force Solution

Approach

The brute force approach to climbing stairs is like exploring every single path you could possibly take to reach the top. It involves systematically checking out each sequence of steps you might make.

Here's how the algorithm would work step-by-step:

Imagine you are at the very bottom of the stairs.
From your current position, consider all the possible next moves you can make, which are either taking one step or taking two steps.
For each of those possible moves, explore all the subsequent moves you could make from the new position.
Continue this process, branching out and exploring every single combination of one-step and two-step moves until you reach the very top.
Each time you successfully reach the top, count it as one valid way to climb the stairs.
After exploring all possible sequences of moves, the total count of ways you reached the top is your answer.

Code Implementation

def climb_stairs_brute_force(number_of_stairs):    def explore_paths(current_stair):        # Base case: if we've reached the top, we found one valid way.        if current_stair == number_of_stairs:            return 1        # Base case: if we've gone past the top, this path is invalid.        if current_stair > number_of_stairs:            return 0        # Recursively explore taking one step and taking two steps.        ways_from_one_step = explore_paths(current_stair + 1)        ways_from_two_steps = explore_paths(current_stair + 2)        # Summing up the ways from both possible next moves.        return ways_from_one_step + ways_from_two_steps    # Start the exploration from the bottom (stair 0).    return explore_paths(0)

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach described involves exploring every possible path to climb the stairs, which is achieved through recursion. At each step, there are two choices: take one step or take two steps. This branching nature leads to an exponential number of recursive calls. The number of operations roughly doubles with each additional step, mirroring the growth of a binary tree. Therefore, the total operations approximate 2 to the power of n, resulting in a time complexity of O(2^n).

Space Complexity

–

Optimal Solution

Approach

To figure out how many distinct ways there are to reach the top, we can realize that the number of ways to reach a certain step depends only on the ways to reach the steps just before it. This pattern allows us to build up the solution from the bottom.

Here's how the algorithm would work step-by-step:

Imagine you want to reach the very first step. There's only one way: take one step.
Now, consider reaching the second step. You can either take two steps at once, or take one step, then another step. That's two ways.
For the third step, you could have arrived from the second step by taking one more step, or from the first step by taking two steps. So, the total ways to reach the third step is the sum of ways to reach the second step and the ways to reach the first step.
This pattern continues. To find the number of ways to reach any given step, you just need to add up the number of ways to reach the step immediately before it and the number of ways to reach the step two steps before it.
By starting with the known ways for the first few steps and repeatedly applying this simple addition rule, you can efficiently calculate the total number of ways to reach any higher step, including the very top one.

Code Implementation

def climb_stairs(total_steps_to_climb):
    if total_steps_to_climb == 1:
        return 1
    # Initialize base cases for the first two steps.
    ways_to_reach_previous_step = 1
    ways_to_reach_two_steps_before = 1

    # Iterate from the third step up to the total number of steps.
    for current_step_number in range(2, total_steps_to_climb + 1):
        # The number of ways to reach the current step is the sum of ways to reach the two preceding steps.
        current_step_ways = ways_to_reach_previous_step + ways_to_reach_two_steps_before
        # Update values for the next iteration.
        ways_to_reach_two_steps_before = ways_to_reach_previous_step
        ways_to_reach_previous_step = current_step_ways

    return ways_to_reach_previous_step

Big(O) Analysis

Time Complexity

O(n) – The solution calculates the number of ways to reach each step from 1 up to n. For each step, it performs a constant number of operations: a simple addition based on the results for the two preceding steps. Since we iterate through each step from 1 to n exactly once, the total number of operations is directly proportional to n. Therefore, the time complexity is O(n).

Space Complexity

O(1) – The provided solution, as described by the pattern recognition and simple addition rule, only requires a constant number of variables to store the number of ways to reach the previous two steps. The input size N does not influence the number of additional variables used. Therefore, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Input n is 1	The base case correctly returns 1, representing the single way to climb one step.
Input n is 2	The base case correctly returns 2, representing the two ways (1+1 or 2) to climb two steps.
Input n is 3	The algorithm should calculate 3 ways (1+1+1, 1+2, 2+1), which is consistent with the Fibonacci sequence pattern.
Large input n leading to integer overflow	Use a data type capable of storing larger numbers, like long long in C++ or Python's arbitrary precision integers, to prevent overflow.
The problem statement guarantees a positive integer n	Therefore, null or zero inputs are not expected and do not require explicit handling according to the problem constraints.
The problem implies a single distinct solution exists for each n	The dynamic programming approach inherently calculates a unique count for each n, ensuring no ambiguity in multiple valid solutions.
Recursive solution without memoization for large n	A purely recursive solution would lead to exponential time complexity and stack overflow for large n, necessitating memoization or an iterative DP approach.
The number of ways grows rapidly like Fibonacci sequence	The efficient iterative dynamic programming solution handles this growth effectively by calculating each step in linear time.

Input n is 1

How to Handle:

The base case correctly returns 1, representing the single way to climb one step.

Input n is 2

How to Handle:

The base case correctly returns 2, representing the two ways (1+1 or 2) to climb two steps.

Input n is 3

How to Handle:

The algorithm should calculate 3 ways (1+1+1, 1+2, 2+1), which is consistent with the Fibonacci sequence pattern.

Large input n leading to integer overflow

How to Handle:

Use a data type capable of storing larger numbers, like long long in C++ or Python's arbitrary precision integers, to prevent overflow.

The problem statement guarantees a positive integer n

How to Handle:

Therefore, null or zero inputs are not expected and do not require explicit handling according to the problem constraints.

The problem implies a single distinct solution exists for each n

How to Handle:

The dynamic programming approach inherently calculates a unique count for each n, ensuring no ambiguity in multiple valid solutions.

Recursive solution without memoization for large n

How to Handle:

A purely recursive solution would lead to exponential time complexity and stack overflow for large n, necessitating memoization or an iterative DP approach.

The number of ways grows rapidly like Fibonacci sequence

How to Handle:

The efficient iterative dynamic programming solution handles this growth effectively by calculating each step in linear time.

Solution

Clarifying Questions

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:

What are the constraints on the value of 'n', specifically its minimum and maximum possible values?
Is 'n' always a positive integer, or should I consider cases where 'n' might be zero or negative?
Are there any specific performance requirements or time/space complexity targets I should aim for?
Is it possible for 'n' to be so large that the number of distinct ways might exceed standard integer types?
Can 'n' be 1? If so, what is the expected output for n=1?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Imagine you are at the very bottom of the stairs.
From your current position, consider all the possible next moves you can make, which are either taking one step or taking two steps.
For each of those possible moves, explore all the subsequent moves you could make from the new position.
Continue this process, branching out and exploring every single combination of one-step and two-step moves until you reach the very top.
Each time you successfully reach the top, count it as one valid way to climb the stairs.
After exploring all possible sequences of moves, the total count of ways you reached the top is your answer.

Code Implementation

def climb_stairs_brute_force(number_of_stairs):    def explore_paths(current_stair):        # Base case: if we've reached the top, we found one valid way.        if current_stair == number_of_stairs:            return 1        # Base case: if we've gone past the top, this path is invalid.        if current_stair > number_of_stairs:            return 0        # Recursively explore taking one step and taking two steps.        ways_from_one_step = explore_paths(current_stair + 1)        ways_from_two_steps = explore_paths(current_stair + 2)        # Summing up the ways from both possible next moves.        return ways_from_one_step + ways_from_two_steps    # Start the exploration from the bottom (stair 0).    return explore_paths(0)

Big(O) Analysis

Time Complexity

Space Complexity

–

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Imagine you want to reach the very first step. There's only one way: take one step.
Now, consider reaching the second step. You can either take two steps at once, or take one step, then another step. That's two ways.
For the third step, you could have arrived from the second step by taking one more step, or from the first step by taking two steps. So, the total ways to reach the third step is the sum of ways to reach the second step and the ways to reach the first step.
This pattern continues. To find the number of ways to reach any given step, you just need to add up the number of ways to reach the step immediately before it and the number of ways to reach the step two steps before it.
By starting with the known ways for the first few steps and repeatedly applying this simple addition rule, you can efficiently calculate the total number of ways to reach any higher step, including the very top one.

Code Implementation

def climb_stairs(total_steps_to_climb):
    if total_steps_to_climb == 1:
        return 1
    # Initialize base cases for the first two steps.
    ways_to_reach_previous_step = 1
    ways_to_reach_two_steps_before = 1

    # Iterate from the third step up to the total number of steps.
    for current_step_number in range(2, total_steps_to_climb + 1):
        # The number of ways to reach the current step is the sum of ways to reach the two preceding steps.
        current_step_ways = ways_to_reach_previous_step + ways_to_reach_two_steps_before
        # Update values for the next iteration.
        ways_to_reach_two_steps_before = ways_to_reach_previous_step
        ways_to_reach_previous_step = current_step_ways

    return ways_to_reach_previous_step

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input n is 1	The base case correctly returns 1, representing the single way to climb one step.
Input n is 2	The base case correctly returns 2, representing the two ways (1+1 or 2) to climb two steps.
Input n is 3	The algorithm should calculate 3 ways (1+1+1, 1+2, 2+1), which is consistent with the Fibonacci sequence pattern.
Large input n leading to integer overflow	Use a data type capable of storing larger numbers, like long long in C++ or Python's arbitrary precision integers, to prevent overflow.
The problem statement guarantees a positive integer n	Therefore, null or zero inputs are not expected and do not require explicit handling according to the problem constraints.
The problem implies a single distinct solution exists for each n	The dynamic programming approach inherently calculates a unique count for each n, ensuring no ambiguity in multiple valid solutions.
Recursive solution without memoization for large n	A purely recursive solution would lead to exponential time complexity and stack overflow for large n, necessitating memoization or an iterative DP approach.
The number of ways grows rapidly like Fibonacci sequence	The efficient iterative dynamic programming solution handles this growth effectively by calculating each step in linear time.

Input n is 1

How to Handle:

The base case correctly returns 1, representing the single way to climb one step.

Input n is 2

How to Handle:

The base case correctly returns 2, representing the two ways (1+1 or 2) to climb two steps.

Input n is 3

How to Handle:

The algorithm should calculate 3 ways (1+1+1, 1+2, 2+1), which is consistent with the Fibonacci sequence pattern.

Large input n leading to integer overflow

How to Handle:

Use a data type capable of storing larger numbers, like long long in C++ or Python's arbitrary precision integers, to prevent overflow.

The problem statement guarantees a positive integer n

How to Handle:

Therefore, null or zero inputs are not expected and do not require explicit handling according to the problem constraints.

The problem implies a single distinct solution exists for each n

How to Handle:

The dynamic programming approach inherently calculates a unique count for each n, ensuring no ambiguity in multiple valid solutions.

Recursive solution without memoization for large n

How to Handle:

A purely recursive solution would lead to exponential time complexity and stack overflow for large n, necessitating memoization or an iterative DP approach.

The number of ways grows rapidly like Fibonacci sequence

How to Handle:

The efficient iterative dynamic programming solution handles this growth effectively by calculating each step in linear time.