Climbing Stairs Problem in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Climbing Stairs Problem?

It is a problem where you count how many distinct ways you can climb to the top of a staircase with n steps, taking 1 or 2 steps at a time.

beginner

What is the base case for the Climbing Stairs Problem when n = 1?

There is only 1 way to climb 1 step: take a single step.

intermediate

How do you express the number of ways to climb n stairs using previous results?

Ways(n) = Ways(n-1) + Ways(n-2). You can reach step n by taking 1 step from n-1 or 2 steps from n-2.

intermediate

What data structure is commonly used to store intermediate results in the Climbing Stairs Problem?

An array or list is used to store the number of ways to reach each step to avoid repeated calculations.

intermediate

Why is the Climbing Stairs Problem similar to the Fibonacci sequence?

Because the number of ways to climb n stairs follows the same pattern as Fibonacci numbers, where each number is the sum of the two before it.

How many ways are there to climb 3 stairs if you can take 1 or 2 steps at a time?

A 5

B 4

C 2

D 3

The ways are: (1,1,1), (1,2), (2,1) — total 3 ways.

What is the recurrence relation for the Climbing Stairs Problem?

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-2)

You can reach step n from step n-1 or n-2, so total ways add up.

What is the number of ways to climb 1 stair?

A 2

B 0

C 1

D 3

Only one way: take one step.

Which programming technique helps avoid repeated calculations in this problem?

A Dynamic programming

B Recursion without memoization

C Brute force

D Sorting

Dynamic programming stores intermediate results to avoid repeats.

If Ways(4) = 5, what is Ways(5)?

A 7

B 8

C 9

D 10

Ways(5) = Ways(4) + Ways(3) = 5 + 3 = 8.

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

Think about how you can use answers for smaller steps to find the answer for bigger steps.

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

Focus on how each step count 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 minimum number of steps to reach the top

B. The number of distinct ways to reach the top

C. The maximum number of steps you can climb at once

D. The total number of steps climbed

Solution

1. Step 1: Understand the problem statement

   The problem asks for the count of different ways to climb n steps taking 1 or 2 steps at a time.

2. Step 2: Identify what is being counted

   We count distinct sequences of 1-step and 2-step moves that sum to n.

3. Final Answer:

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

4. Quick Check:

   Counting ways = number of sequences [OK]

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

Common Mistakes:

• Confusing number of ways with minimum steps
• Thinking about maximum steps instead of counting ways
• Counting total steps instead of sequences

2. Which of the following C# method signatures correctly represents a function to solve the Climbing Stairs problem for an integer n?

easy

A. public int ClimbStairs(int n)

B. int climb_stairs(string n)

C. void ClimbStairs(int n)

D. public string ClimbStairs(int n)

Solution

1. Step 1: Check method return type and parameters

   The function should return an integer count and take an integer n as input.

2. Step 2: Validate method signature syntax

   public int ClimbStairs(int n) uses public access, returns int, and takes int n, which is correct C# syntax.

3. Final Answer:

   public int ClimbStairs(int n) -> Option A

4. Quick Check:

   Correct return type and parameter [OK]

Hint: Return int, input int n, use public int method [OK]

Common Mistakes:

• Using void return type instead of int
• Using string parameter instead of int
• Returning string instead of int

3. What is the output of the following C# code for n = 4?

public int ClimbStairs(int n) {
    if (n <= 2) return n;
    int a = 1, b = 2, c = 0;
    for (int i = 3; i <= n; i++) {
        c = a + b;
        a = b;
        b = c;
    }
    return c;
}

Console.WriteLine(ClimbStairs(4));

medium

A. 5

B. 6

C. 7

D. 4

Solution

1. Step 1: Trace the loop for n=4

   Start with a=1, b=2. For i=3: c=1+2=3, a=2, b=3. For i=4: c=2+3=5, a=3, b=5.

2. Step 2: Return value after loop

   After loop ends, c=5 is returned and printed.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Ways for 4 steps = 5 [OK]

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

Common Mistakes:

• Returning b instead of c
• Starting loop from wrong index
• Incorrect initial values for a and b

4. Identify the error in this C# code snippet for climbing stairs:

public int ClimbStairs(int n) {
    if (n == 1) return 1;
    int[] dp = new int[n];
    dp[0] = 1;
    dp[1] = 2;
    for (int i = 2; i <= n; i++) {
        dp[i] = dp[i-1] + dp[i-2];
    }
    return dp[n];
}

medium

A. Loop should start from i = 1

B. Incorrect base case for n == 1

C. Wrong return statement should be dp[n-1]

D. Array index out of bounds error due to dp size

Solution

1. Step 1: Check array size and indexing

   Array dp has size n, valid indices 0 to n-1. Loop runs i to n inclusive, accessing dp[n] which is out of bounds.

2. Step 2: Identify cause of error

   Accessing dp[n] causes runtime error. Loop should run to i < n, and return dp[n-1].

3. Final Answer:

   Array index out of bounds error due to dp size -> Option D

4. Quick Check:

   dp array size mismatch causes error [OK]

Hint: Array size n means last index n-1, not n [OK]

Common Mistakes:

• Loop running to i <= n instead of i < n
• Returning dp[n] instead of dp[n-1]
• Not initializing dp array correctly

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 approach correctly adapts the Climbing Stairs problem?

hard

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

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

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

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 i depend on previous 3 dp values.

2. Step 2: Formulate dp relation and base cases

   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 C

4. Quick Check:

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

Hint: Add last 3 dp values for steps 1,2,3 [OK]

Common Mistakes:

• Multiplying dp values instead of adding
• Using only last 2 dp values
• Not setting correct base cases