Climbing Stairs Problem in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind the Climbing Stairs Problem?

It counts how many different ways you can climb to the top of a staircase if you can take either 1 or 2 steps at a time.

intermediate

How can the Climbing Stairs Problem be related to the Fibonacci sequence?

The number of ways to reach step n equals the sum of ways to reach steps n-1 and n-2, just like Fibonacci numbers.

beginner

What is the base case for the Climbing Stairs Problem?

For step 0 (ground), there is 1 way (do nothing), and for step 1, there is 1 way (one step).

intermediate

Why is dynamic programming useful for the Climbing Stairs Problem?

It saves results of smaller steps to avoid repeating calculations, making the solution faster.

intermediate

What is the time complexity of the dynamic programming solution for the Climbing Stairs Problem?

O(n), where n is the number of steps, because each step is computed once.

In the Climbing Stairs Problem, how many ways are there to climb 3 steps?

A 4

B 3

C 2

D 5

Ways to climb 3 steps: (1+1+1), (1+2), (2+1) = 3 ways.

Which of these best describes the relation between ways to climb step n and previous steps?

A ways(n) = ways(n-1) - ways(n-2)

B ways(n) = ways(n-1) * ways(n-2)

C ways(n) = ways(n-1) + ways(n-2)

D ways(n) = ways(n-1) / ways(n-2)

The number of ways to reach step n is the sum of ways to reach n-1 and n-2.

What is the minimum number of steps you can take at once in this problem?

A 1

B 2

C 3

D 0

You can take either 1 or 2 steps at a time, so minimum is 1.

What technique helps avoid repeated calculations in the Climbing Stairs Problem?

A Dynamic Programming

B Brute Force

C Sorting

D Greedy Algorithm

Dynamic Programming stores results of smaller problems to reuse them.

If you have 0 steps, how many ways are there to climb?

A 0

B Undefined

C 2

D 1

There is 1 way: do nothing (stay at the bottom).

Explain how to solve the Climbing Stairs Problem using dynamic programming.

Think about how you can build the answer step by step.

Describe why the Climbing Stairs Problem is similar to the Fibonacci sequence.

Focus on how each step depends on the two before it.

Practice

1. What does the Climbing Stairs problem ask you to find?
Given n steps, how many distinct ways can you climb to the top if you can take 1 or 2 steps at a time?

easy

A. The maximum number of steps you can skip

B. The total number of distinct ways to reach the top

C. The minimum number of steps to reach the top

D. The number of ways to climb only 1 step at a time

Solution

1. Step 1: Understand the problem statement

   The problem asks for the count of different ways to reach the top by taking either 1 or 2 steps at a time.

2. Step 2: Identify what is being counted

   We count all possible sequences of 1-step and 2-step moves that sum to n steps.

3. Final Answer:

   The total number of distinct ways to reach the top -> Option B

4. Quick Check:

   Counting distinct sequences = total ways [OK]

Hint: Count sequences of 1 or 2 steps summing to n [OK]

Common Mistakes:

• Confusing number of ways with minimum steps
• Counting only 1-step moves
• Ignoring 2-step moves

2. Which of the following Rust function signatures correctly represents a function to solve the Climbing Stairs problem for input n: i32 returning an i32?

easy

A. fn climb_stairs(n: i32) -> String

B. fn climb_stairs(n: i32, result: i32)

C. fn climb_stairs(n: String) -> i32

D. fn climb_stairs(n: i32) -> i32

Solution

1. Step 1: Check parameter and return types

   The input is number of steps as i32, and output is number of ways as i32.

2. Step 2: Match function signature

   Only fn climb_stairs(n: i32) -> i32 matches input and output types correctly.

3. Final Answer:

   fn climb_stairs(n: i32) -> i32 -> Option D

4. Quick Check:

   Input and output types match function signature [OK]

Hint: Input and output types must match problem data [OK]

Common Mistakes:

• Using wrong return type like String
• Passing input as String instead of i32
• Missing return type or extra parameters

3. What is the output of this Rust code for n = 4?

fn climb_stairs(n: i32) -> i32 {
    if n <= 2 {
        return n;
    }
    let mut first = 1;
    let mut second = 2;
    let mut ways = 0;
    for _ in 3..=n {
        ways = first + second;
        first = second;
        second = ways;
    }
    ways
}

fn main() {
    println!("{}", climb_stairs(4));
}

medium

A. 4

B. 6

C. 5

D. 7

Solution

1. Step 1: Trace initial values and loop

   n=4, first=1, second=2. Loop runs for i=3 and i=4. - i=3: ways=1+2=3, first=2, second=3 - i=4: ways=2+3=5, first=3, second=5

2. Step 2: Return final ways value

   After loop, ways=5 is returned and printed.

3. Final Answer:

   5 -> Option C

4. Quick Check:

   Ways for 4 steps = 5 [OK]

Hint: Use Fibonacci-like addition for steps count [OK]

Common Mistakes:

• Returning wrong initial value
• Off-by-one errors in loop
• Mixing first and second variables

4. Identify the bug in this Rust code for climbing stairs and select the fix:

fn climb_stairs(n: i32) -> i32 {
    if n == 1 {
        return 1;
    }
    let mut first = 1;
    let mut second = 2;
    let mut ways = 0;
    for i in 2..n {
        ways = first + second;
        first = second;
        second = ways;
    }
    ways
}

medium

A. Return second instead of ways at the end

B. Initialize ways to 1 instead of 0

C. Change if condition to n <= 2

D. Change loop to for i in 3..=n

Solution

1. Step 1: Identify the issue with base case n=2

   For n=2, if not triggered, loop 2..2 empty, returns ways=0 (wrong, should be 2). second initialized to 2.

2. Step 2: Observe the update logic

   Each loop sets second to the new ways value (nth ways). After loop (or none), second is correct: initial 2 for n=2, updated for larger n.

3. Final Answer:

   Return second instead of ways at the end -> Option A

4. Quick Check:

   second holds nth ways after shifts [OK]

Hint: second tracks the current (nth) ways value [OK]

Common Mistakes:

• Loop runs too short or too long
• Wrong initial if condition
• Returning wrong variable

5. You want to find the number of ways to climb 10 steps but now you can take 1, 2, or 3 steps at a time. Which change to the climbing stairs logic is correct to handle this?

hard

A. Use dp[i] = dp[i-1] + dp[i-2] + dp[i-3] with base cases dp[0]=1, dp[1]=1, dp[2]=2

B. Use dp[i] = dp[i-1] * dp[i-2] * dp[i-3] with base cases dp[0]=1, dp[1]=1, dp[2]=2

C. Use dp[i] = dp[i-1] + dp[i-2] only, ignore 3 steps

D. Use recursion without memoization for all steps

Solution

1. Step 1: Understand the new step options

   Now steps can be 1, 2, or 3 at a time, so ways to reach step i depend on previous 3 steps.

2. Step 2: Update recurrence relation and base cases

   Use dp[i] = dp[i-1] + dp[i-2] + dp[i-3]. Base cases: dp[0]=1 (one way to stand still), dp[1]=1, dp[2]=2.

3. Final Answer:

   Use dp[i] = dp[i-1] + dp[i-2] + dp[i-3] with base cases dp[0]=1, dp[1]=1, dp[2]=2 -> Option A

4. Quick Check:

   Sum of last 3 dp values for each step [OK]

Hint: Sum ways from last 3 steps for new step sizes [OK]

Common Mistakes:

• Multiplying dp values instead of adding
• Ignoring 3-step moves
• Using recursion without memoization causing slowness