Why Climbing Stairs Problem in DSA C++?

The Big Idea

What if you could count all ways to climb stairs without writing a single path?

The Scenario

Imagine you want to climb a staircase with many steps. You try to count all the ways to reach the top by writing each path on paper.

For example, you write down: step 1, step 2, step 1+1, step 2+1, and so on.

The Problem

Writing all possible step combinations manually is slow and confusing.

As the number of steps grows, the list becomes huge and hard to manage.

You might miss some combinations or count some twice.

The Solution

The Climbing Stairs Problem uses a simple rule: the ways to reach step n equals the ways to reach step n-1 plus the ways to reach step n-2.

This rule helps calculate the total ways quickly without listing all paths.

Before vs After

✗ Before

int countWays(int steps) {
  // Manually list all paths - very long and complex
  // Not practical for large steps
  return 0; // placeholder
}

✓ After

int countWays(int steps) {
  if (steps <= 2) return steps;
  int waysOneStepBefore = 2;
  int waysTwoStepsBefore = 1;
  int totalWays = 0;
  for (int currentStep = 3; currentStep <= steps; currentStep++) {
    totalWays = waysOneStepBefore + waysTwoStepsBefore;
    waysTwoStepsBefore = waysOneStepBefore;
    waysOneStepBefore = totalWays;
  }
  return totalWays;
}

What It Enables

This concept lets you find the number of ways to climb any staircase quickly and without errors.

Real Life Example

Planning how many ways a child can climb stairs taking one or two steps at a time, helping design safe playground steps.

Key Takeaways

Manual counting of step combinations is slow and error-prone.

The problem uses a simple addition rule to find total ways efficiently.

This approach scales well for large numbers of 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