Climbing Stairs Problem in DSA C#

Mental Model

You want to find how many ways you can reach the top of stairs by taking 1 or 2 steps at a time.

Analogy: Imagine climbing a staircase where you can take either one step or two steps each time. Counting all possible ways to reach the top is like counting all different paths you can take.

Start: 0
Steps: 1 2 3 4 5 ... n
Each step can be reached from the previous step or two steps before.

Dry Run Walkthrough

Input: stairs: 4 steps

Goal: Find how many distinct ways to reach step 4 by taking 1 or 2 steps at a time

Step 1: Initialize ways to reach step 1 and step 2

ways[1] = 1, ways[2] = 2

Why: Base cases: 1 way to reach step 1 (one step), 2 ways to reach step 2 (1+1 or 2)

Step 2: Calculate ways to reach step 3 as sum of ways to reach steps 2 and 1

ways[3] = ways[2] + ways[1] = 2 + 1 = 3

Why: You can get to step 3 from step 2 or step 1

Step 3: Calculate ways to reach step 4 as sum of ways to reach steps 3 and 2

ways[4] = ways[3] + ways[2] = 3 + 2 = 5

Why: You can get to step 4 from step 3 or step 2

Result:

ways array: [1, 2, 3, 5]
Answer: 5 ways to climb 4 steps

Annotated Code

DSA C#

using System;

class ClimbingStairs {
    public static int CountWays(int n) {
        if (n == 0) return 1; // 1 way to stand still at the bottom
        if (n == 1) return 1; // 1 way to climb 1 step

        int[] ways = new int[n + 1];
        ways[1] = 1; // base case: 1 way to reach step 1
        ways[2] = 2; // base case: 2 ways to reach step 2

        for (int i = 3; i <= n; i++) {
            ways[i] = ways[i - 1] + ways[i - 2]; // sum ways to reach previous two steps
        }

        return ways[n];
    }

    static void Main() {
        int stairs = 4;
        int result = CountWays(stairs);
        Console.WriteLine(result);
    }
}

if (n == 0) return 1; // 1 way to stand still at the bottom

handle base case where stairs are 0

if (n == 1) return 1; // 1 way to climb 1 step

handle base case where stairs are 1

ways[1] = 1; // base case: 1 way to reach step 1

initialize ways to reach first step

ways[2] = 2; // base case: 2 ways to reach step 2

initialize ways to reach second step

for (int i = 3; i <= n; i++) { ways[i] = ways[i - 1] + ways[i - 2]; // sum ways to reach previous two steps }

calculate ways to reach each step by summing ways to previous two steps

return ways[n];

return total ways to reach nth step

Output

5

Complexity Analysis

Time: O(n) because we calculate ways for each step from 1 to n once

Space: O(n) because we store ways for all steps up to n in an array

vs Alternative: Compared to recursive approach without memoization which is exponential, this dynamic programming approach is efficient and avoids repeated calculations

Edge Cases

n = 0 (no steps)

Returns 1 because there is one way to stand still at the bottom

DSA C#

if (n == 0) return 1; // 1 way to stand still at the bottom

n = 1 (one step)

Returns 1 because only one step can be taken

DSA C#

if (n == 1) return 1; // 1 way to climb 1 step

n = 2 (two steps)

Returns 2 because you can take two single steps or one double step

DSA C#

ways[2] = 2; // base case: 2 ways to reach step 2

When to Use This Pattern

When a problem asks for counting ways to reach a point with fixed step sizes, use dynamic programming summing previous results because each step depends on the last two.

Common Mistakes

Mistake: Not handling base cases correctly, causing index errors or wrong results
Fix: Explicitly return 1 for n == 0 and n == 1 and initialize ways[1] and ways[2] properly

Mistake: Using recursion without memoization leading to slow exponential time
Fix: Use bottom-up dynamic programming with an array to store intermediate results

Summary

Counts the number of ways to climb n stairs taking 1 or 2 steps at a time.

Use when you need to count distinct paths with fixed step increments.

Each step's ways count is the sum of ways to the two previous steps.

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