Climbing Stairs Problem in DSA C# - Execution Trace

Concept Flow - Climbing Stairs Problem

Start at step 0

↓

Can climb 1 or 2 steps

↓

Calculate ways to reach step i

↓

ways[i

↓

Repeat until step n

↓

Return ways[n

We start at step 0 and calculate the number of ways to reach each step by adding ways to reach the previous two steps until we reach step n.

Execution Sample

DSA C#

int ClimbStairs(int n) {
  if (n < 2) return 1;
  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];
  }
  return ways[n];
}

This code calculates the number of ways to climb n stairs by summing ways to reach the previous two steps.

Execution Table

Step	Operation	Index i	ways[i-2]	ways[i-1]	ways[i]	Visual State of ways array
1	Initialize base cases	N/A	N/A	N/A	N/A	[1, 1, 0, 0, 0, ...]
2	Calculate ways[2]	2	ways[0]=1	ways[1]=1	2	[1, 1, 2, 0, 0, ...]
3	Calculate ways[3]	3	ways[1]=1	ways[2]=2	3	[1, 1, 2, 3, 0, ...]
4	Calculate ways[4]	4	ways[2]=2	ways[3]=3	5	[1, 1, 2, 3, 5, ...]
5	Calculate ways[5]	5	ways[3]=3	ways[4]=5	8	[1, 1, 2, 3, 5, 8]

💡 Reached step n=5, calculation complete.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	Final
i	N/A	2	3	4	5	5
ways[0]	1	1	1	1	1	1
ways[1]	1	1	1	1	1	1
ways[2]	0	2	2	2	2	2
ways[3]	0	0	3	3	3	3
ways[4]	0	0	0	5	5	5
ways[5]	0	0	0	0	8	8

Key Moments - 3 Insights

Why do we set ways[0] and ways[1] to 1 at the start?

Because ways[0] represents the base case of standing at the bottom (1 way), and ways[1] means there is only 1 way to reach the first step (one single step). This is shown in execution_table row 1.

Why do we add ways[i-1] and ways[i-2] to get ways[i]?

Because to reach step i, you can come from step i-1 by taking 1 step or from step i-2 by taking 2 steps. So total ways to i is sum of ways to i-1 and i-2, as shown in rows 2 to 5.

What happens if n is 1 or 0?

If n is 0 or 1, the function returns 1 immediately because there's only one way to stand still or take one step, matching the base cases in the code and execution_table row 1.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the value of ways[3]?

A 2

B 5

C 3

D 1

💡 Hint

Check the 'ways[i]' column in execution_table row 3.

At which step does the ways array first contain the value 5?

A Step 3

B Step 4

C Step 5

D Step 2

💡 Hint

Look at the 'Visual State of ways array' column in execution_table rows.

If ways[0] was set to 0 instead of 1, how would ways[2] change?

A ways[2] would be 1

B ways[2] would be 0

C ways[2] would be 2

D ways[2] would be 3

💡 Hint

ways[2] = ways[1] + ways[0], check variable_tracker for ways[0] and ways[1].

Concept Snapshot

Climbing Stairs Problem:
- ways[0] = 1, ways[1] = 1 base cases
- For i >= 2: ways[i] = ways[i-1] + ways[i-2]
- Represents number of ways to reach step i
- Use array to store intermediate results
- Return ways[n] as final answer

Full Transcript

The Climbing Stairs Problem counts how many ways you can climb to the top of n stairs by taking 1 or 2 steps at a time. We start with base cases: 1 way to stand at step 0 and 1 way to reach step 1. For each step from 2 to n, the number of ways to reach it is the sum of ways to reach the previous step and the step before that. This is because you can get to step i by either taking one step from i-1 or two steps from i-2. We store these counts in an array called ways. The code initializes ways[0] and ways[1] to 1, then fills the array up to ways[n]. The final answer is ways[n]. The execution table shows how the array fills step by step, and the variable tracker shows how variables change. Key moments clarify why base cases are set and why we add the two previous counts. The quiz tests understanding of these steps and values.

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