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 either 1 or 2 steps at a time.

beginner

What is the base case for the Climbing Stairs Problem?

For 1 step, there is 1 way to climb. For 2 steps, there are 2 ways: (1+1) or (2).

intermediate

Which mathematical sequence does the Climbing Stairs Problem relate to?

It relates to the Fibonacci sequence because the number of ways to climb n steps equals the sum of ways to climb (n-1) and (n-2) steps.

intermediate

What is the recurrence relation for the Climbing Stairs Problem?

ways(n) = ways(n-1) + ways(n-2), where ways(n) is the number of ways to climb n steps.

intermediate

Why is dynamic programming useful for the Climbing Stairs Problem?

Dynamic programming stores results of smaller problems (like ways to climb fewer steps) to avoid repeated calculations, making the solution efficient.

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

A 3

B 2

C 4

D 5

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

What does ways(n) depend on in the Climbing Stairs Problem?

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

B ways(n-1) only

C ways(n-2) only

D ways(n-3) and ways(n-4)

ways(n) = ways(n-1) + ways(n-2) because you can reach step n from step n-1 or n-2.

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

A O(n^2)

B O(n)

C O(2^n)

D O(log n)

Dynamic programming solves the problem in linear time by storing intermediate results.

Which of these is NOT a valid step size in the classic Climbing Stairs Problem?

A 1

B 2

C 3

D Both 1 and 2

The classic problem only allows steps of size 1 or 2.

If ways(1) = 1 and ways(2) = 2, what is ways(4)?

A 6

B 8

C 7

D 5

ways(3) = 3, ways(4) = ways(3) + ways(2) = 3 + 2 = 5.

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

Think about how you can build the solution step by step using previous answers.

Describe the relationship between the Climbing Stairs Problem and the Fibonacci sequence.

Focus on how the number of ways to climb relates to previous counts.

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 take at once

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 steps you skip

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 calculated

   We want the total number of distinct sequences of 1-step and 2-step moves that sum to n.

3. Final Answer:

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

4. Quick Check:

   Count ways to climb = total distinct 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
• Ignoring the step size constraints

2. Which of the following is the correct C++ function signature to solve the Climbing Stairs problem recursively?

easy

A. int climbStairs(int &n);

B. void climbStairs(int n);

C. int climbStairs(int n);

D. int climbStairs();

Solution

1. Step 1: Identify the return type

   The function should return an integer count of ways, so return type is int.

2. Step 2: Check the parameter list

   We need the number of steps n as input, passed by value as int n.

3. Final Answer:

   int climbStairs(int n); -> Option C

4. Quick Check:

   Return type int and parameter int n [OK]

Hint: Function returns int count and takes int n parameter [OK]

Common Mistakes:

• Using void return type instead of int
• Missing the parameter n
• Passing n by reference unnecessarily

3. What is the output of the following C++ code?

int climbStairs(int n) {
  if (n <= 2) return n;
  return climbStairs(n-1) + climbStairs(n-2);
}

int main() {
  std::cout << climbStairs(4);
  return 0;
}

medium

A. 7

B. 3

C. 4

D. 5

Solution

1. Step 1: Trace climbStairs(4) calls

   climbStairs(4) = climbStairs(3) + climbStairs(2) climbStairs(3) = climbStairs(2) + climbStairs(1) = 2 + 1 = 3 climbStairs(2) = 2

2. Step 2: Calculate final result

   climbStairs(4) = 3 + 2 = 5

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Ways for 4 steps = 5 [OK]

Hint: Use recursion formula: ways(n) = ways(n-1) + ways(n-2) [OK]

Common Mistakes:

• Returning n instead of sum for n > 2
• Miscounting base cases
• Confusing return values for climbStairs(3)

4. Find the error in this C++ code for climbing stairs:

int climbStairs(int n) {
  if (n == 1) return 1;
  if (n == 2) return 2;
  return climbStairs(n-1) * climbStairs(n-2);
}

medium

A. Using multiplication instead of addition in recursion

B. Base cases are incorrect

C. Missing return type

D. Function parameter should be passed by reference

Solution

1. Step 1: Check base cases

   Base cases for n=1 and n=2 are correct returning 1 and 2 respectively.

2. Step 2: Analyze recursive return statement

   It uses multiplication (*) instead of addition (+), which is wrong for counting ways.

3. Final Answer:

   Using multiplication instead of addition in recursion -> Option A

4. Quick Check:

   Recursion sums ways, not multiplies [OK]

Hint: Sum ways from previous two steps, don't multiply [OK]

Common Mistakes:

• Multiplying instead of adding recursive calls
• Incorrect base cases for n=1 or n=2
• Passing parameter by reference unnecessarily

5. You want to find the number of ways to climb 10 steps but can take 1, 2, or 3 steps at a time. Which approach correctly adapts the Climbing Stairs problem?

hard

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

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

C. Use recursion with only dp[n] = dp[n-1] + dp[n-2]

D. Use dp[n] = dp[n-1] + dp[n-3] ignoring dp[n-2]

Solution

1. Step 1: Understand the step options

   Since steps can be 1, 2, or 3, ways to reach step n = sum of ways to reach n-1, n-2, and n-3.

2. Step 2: Define base cases and recurrence

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

3. Final Answer:

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

4. Quick Check:

   Sum ways from last 3 steps with correct bases [OK]

Hint: Add ways from last 3 steps for 1,2,3 step moves [OK]

Common Mistakes:

• Multiplying ways instead of adding
• Ignoring dp[n-3] term
• Wrong base cases causing incorrect counts