Climbing Stairs Problem in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the base case for 1 stair.

DSA C++

if (n == 1) return [1];

Drag options to blanks, or click blank then click option'

A0

Bn

C1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 instead of 1 for the base case.

Returning n which is unnecessary for base case.

When there is only 1 stair, there is exactly 1 way to climb it.

2fill in blank

medium

Complete the code to return the base case for 2 stairs.

DSA C++

if (n == 2) return [1];

Drag options to blanks, or click blank then click option'

A0

B1

C3

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 1 instead of 2 for 2 stairs.

Confusing the number of steps with the number of ways.

For 2 stairs, you can climb either 1 step + 1 step or directly 2 steps, so 2 ways.

3fill in blank

hard

Fix the error in the recursive return statement to calculate ways correctly.

DSA C++

return [1](n - 1) + climbStairs(n - 2);

Drag options to blanks, or click blank then click option'

AclimbStairs

Bclimbstairs

CclimbStair

Dclimb_Stairs

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect capitalization in function name.

Typos in function name causing compilation errors.

The function name must match exactly, including case: climbStairs.

4fill in blank

hard

Fill both blanks to complete the memoization check and store the result.

DSA C++

if (memo[[1]] != -1) return memo[[1]];
memo[[2]] = climbStairs(n - 1) + climbStairs(n - 2);

Drag options to blanks, or click blank then click option'

An

Bn - 1

Cn - 2

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using different indices for checking and storing memo results.

Using indices like n-1 or n-2 incorrectly for memo storage.

We check and store the result for the current stair number n in memo[n].

5fill in blank

hard

Fill all three blanks to create a bottom-up dynamic programming solution.

DSA C++

vector<int> dp(n + 1, 0);
dp[0] = 1;
dp[1] = 1;
for (int i = [1]; i <= n; i++) {
    dp[i] = dp[i [2] 1] + dp[i [3] 2];
}
return dp[n];

Drag options to blanks, or click blank then click option'

A2

B-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop from 1 instead of 2.

Using addition instead of subtraction for indices.

Using subtraction instead of addition for combining ways.

Start loop from 2; use subtraction to access previous steps; add ways from 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 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