Climbing Stairs Problem in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Climbing Stairs Recursive Code

What is the output of the following C++ code that calculates the number of ways to climb 4 stairs using recursion?

DSA C++

#include <iostream>
using namespace std;

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

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

A 6

B 7

C 5

D 4

💡 Hint

Think about how many ways to reach step 4 by summing ways to reach step 3 and step 2.

The function climbStairs uses the Fibonacci pattern. For n=4, ways = climbStairs(3) + climbStairs(2) = 3 + 2 = 5.

❓ Predict Output

intermediate

Output of Climbing Stairs with Memoization

What is the output of this C++ code that uses memoization to calculate ways to climb 5 stairs?

DSA C++

#include <iostream>
#include <vector>
using namespace std;

int climbStairsMemo(int n, vector<int>& memo) {
    if (n <= 2) return n;
    if (memo[n] != -1) return memo[n];
    memo[n] = climbStairsMemo(n - 1, memo) + climbStairsMemo(n - 2, memo);
    return memo[n];
}

int main() {
    int n = 5;
    vector<int> memo(n + 1, -1);
    cout << climbStairsMemo(n, memo) << endl;
    return 0;
}

A 7

B 10

C 5

D 8

💡 Hint

Memoization stores results to avoid repeated calculations.

For n=5, ways = climbStairs(4) + climbStairs(3) = 5 + 3 = 8.

🔧 Debug

advanced

Identify the Error in Iterative Climbing Stairs Code

What error will this C++ code produce when trying to calculate the number of ways to climb 3 stairs iteratively?

DSA C++

#include <iostream>
using namespace std;

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

int main() {
    cout << climbStairs(3) << endl;
    return 0;
}

A Runtime error: Array index out of bounds

B Output: 5

C Output: 3

D Compilation error: dp array size mismatch

💡 Hint

Check the size of the dp array and the loop index.

The dp array has size 3 but the loop tries to access dp[3] when i=3, causing out of bounds access.

🧠 Conceptual

advanced

Number of Ways to Climb Stairs with Steps 1, 2, or 3

If you can climb stairs by taking 1, 2, or 3 steps at a time, how many ways are there to climb 4 stairs?

A 7

B 6

C 4

D 8

💡 Hint

Count all sequences of 1, 2, and 3 steps that sum to 4.

Ways to climb 4 stairs with steps 1,2,3 are: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2, 1+3, 3+1 totaling 7 ways.

🚀 Application

expert

Minimum Steps to Reach Top with Variable Step Sizes

Given an array steps = {1, 3, 5}, representing allowed step sizes, what is the minimum number of steps to climb 11 stairs?

A 4

B 3

C 5

D 6

💡 Hint

Try to use the largest steps first and check combinations.

One minimum way: 5 + 5 + 1 = 3 steps is valid because 5+5=10 +1=11 with 3 steps total.

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