Climbing Stairs Problem in DSA C# - Build from Scratch

Climbing Stairs Problem

📖 Scenario: Imagine you are climbing a staircase that has a certain number of steps. You can climb either 1 step or 2 steps at a time. You want to find out how many different ways you can reach the top.

🎯 Goal: Build a program that calculates the number of distinct ways to climb to the top of a staircase with n steps, where you can take either 1 or 2 steps at a time.

📋 What You'll Learn

Create an integer variable n representing the total number of steps in the staircase.

Create an integer array ways to store the number of ways to reach each step.

Use a loop to fill the ways array based on the climbing rules.

Print the total number of ways to reach the top step.

💡 Why This Matters

🌍 Real World

This problem models situations where you have multiple ways to reach a goal by taking small steps, like climbing stairs or moving through stages.

💼 Career

Understanding this problem helps with dynamic programming concepts, which are useful in software engineering roles that involve optimization and algorithm design.

1

Setup the number of steps

Create an integer variable called n and set it to 5 to represent the total number of steps in the staircase.

DSA C#

// Your code here

Hint

Use int n = 5; to create the variable.

2

Create an array to store ways

Create an integer array called ways with size n + 1 to store the number of ways to reach each step. Initialize ways[0] to 1 and ways[1] to 1.

DSA C#

int n = 5;
// Your code here

Hint

Use int[] ways = new int[n + 1]; and set the first two elements to 1.

3

Calculate ways to climb

Use a for loop with variable i starting from 2 up to and including n. Inside the loop, set ways[i] to ways[i - 1] + ways[i - 2].

DSA C#

int n = 5;
int[] ways = new int[n + 1];
ways[0] = 1;
ways[1] = 1;
// Your code here

Hint

Use a loop from 2 to n and update ways[i] as the sum of the two previous values.

4

Print the total ways

Print the value of ways[n] using Console.WriteLine to display the total number of ways to climb the staircase.

DSA C#

int n = 5;
int[] ways = new int[n + 1];
ways[0] = 1;
ways[1] = 1;
for (int i = 2; i <= n; i++)
{
    ways[i] = ways[i - 1] + ways[i - 2];
}
// Your code here

Hint

Use Console.WriteLine(ways[n]); to print the result.

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