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 ways you can climb to the top of a staircase if you can take either 1 or 2 steps at a time.

beginner

What is the base case for the Climbing Stairs Problem?

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

intermediate

How do you calculate the number of ways to climb n stairs?

The number of ways to climb n stairs equals the sum of ways to climb (n-1) stairs and ways to climb (n-2) stairs.

intermediate

Why is the Climbing Stairs Problem similar to the Fibonacci sequence?

Because each number of ways depends on the sum of the two previous counts, just like Fibonacci numbers.

advanced

What is the time complexity of the simple recursive solution for the Climbing Stairs Problem?

It is exponential (O(2^n)) because it recalculates the same values many times.

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

A 5

B 4

C 2

D 3

The ways are: (1+1+1), (1+2), (2+1) which totals 3.

What is the recurrence relation for the Climbing Stairs Problem?

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

B ways(n) = ways(n-1) * ways(n-2)

C ways(n) = ways(n-1) + ways(n-2)

D ways(n) = ways(n-1) / ways(n-2)

The number of ways to climb n stairs is the sum of ways to climb n-1 and n-2 stairs.

What is the simplest base case for the Climbing Stairs Problem?

A ways(1) = 1

B ways(2) = 3

C ways(0) = 1

D ways(3) = 2

There is only 1 way to climb 1 stair: a single step.

Which approach improves the time complexity of the Climbing Stairs Problem from exponential to linear?

A Memoization

B Brute force recursion

C Random guessing

D Sorting

Memoization stores results of subproblems to avoid repeated calculations.

If you can climb 1, 2, or 3 steps at a time, how does the recurrence relation change?

A ways(n) = ways(n-1) * ways(n-2) * ways(n-3)

B ways(n) = ways(n-1) + ways(n-2) + ways(n-3)

C ways(n) = ways(n-1) - ways(n-2) - ways(n-3)

D ways(n) = ways(n-1) / ways(n-2) / ways(n-3)

You add the ways to climb n-1, n-2, and n-3 stairs because you can take 1, 2, or 3 steps.

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

Think about how you can use answers for smaller stairs to find answers for bigger stairs.

Describe why the Climbing Stairs Problem is related to the Fibonacci sequence.

Focus on how the count for step n depends on counts for steps n-1 and n-2.

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 number of steps you skip

B. The total number of distinct ways to reach the top

C. The maximum number of steps you can take at once

D. The minimum number of steps to reach the top

Solution

1. Step 1: Understand the problem statement

   The problem asks for the count of all possible ways to climb to the top using steps of size 1 or 2.

2. Step 2: Identify what is being counted

   We count distinct sequences of 1-step and 2-step moves that sum to n steps.

3. Final Answer:

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

4. Quick Check:

   Counting ways = 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 step size instead of counting ways
• Ignoring the step size options

2. Which of the following is the correct base case initialization for the Climbing Stairs problem in C?

easy

A. int ways[1] = 1; int ways[2] = 2;

B. int ways[0] = 1; int ways[1] = 2;

C. int ways[0] = 0; int ways[1] = 1;

D. int ways[1] = 0; int ways[2] = 1;

Solution

1. Step 1: Identify base cases for climbing stairs

   For 1 step, ways = 1 (only one way). For 2 steps, ways = 2 (1+1 or 2).

2. Step 2: Match base cases with options

   int ways[1] = 1; int ways[2] = 2; correctly sets ways[1] = 1 and ways[2] = 2.

3. Final Answer:

   int ways[1] = 1; int ways[2] = 2; -> Option A

4. Quick Check:

   Base cases: ways[1]=1, ways[2]=2 [OK]

Hint: Remember ways(1)=1 and ways(2)=2 base cases [OK]

Common Mistakes:

• Setting ways[0] incorrectly
• Confusing ways for 1 and 2 steps
• Using zero for ways[1]

3. What is the output of the following C code snippet for n = 4?

int climbStairs(int n) {
    if (n <= 2) return n;
    int a = 1, b = 2, c;
    for (int i = 3; i <= n; i++) {
        c = a + b;
        a = b;
        b = c;
    }
    return b;
}

printf("%d", climbStairs(4));

medium

A. 7

B. 6

C. 5

D. 4

Solution

1. Step 1: Trace the loop for n=4

   Initial: a=1, b=2 i=3: c=1+2=3, a=2, b=3 i=4: c=2+3=5, a=3, b=5

2. Step 2: Return value after loop

   Return b=5, which is the number of ways to climb 4 steps.

3. Final Answer:

   5 -> Option C

4. Quick Check:

   Ways(4) = 5 [OK]

Hint: Use Fibonacci-like addition for steps > 2 [OK]

Common Mistakes:

• Returning a instead of b
• Off-by-one errors in loop
• Incorrect initial values

4. Identify the bug in this C function for climbing stairs:

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

medium

A. Array ways is not initialized for ways[0]

B. No bug, code works correctly

C. Accessing ways[2] before initialization

D. Loop should start from i = 2

Solution

1. Step 1: Check array initialization

   Array ways is declared but ways[0] is not initialized, which may cause undefined behavior if accessed.

2. Step 2: Analyze usage of ways[0]

   Though ways[0] is not used in the loop, some compilers or logic might expect it initialized; best practice is to initialize ways[0] = 0.

3. Final Answer:

   Array ways is not initialized for ways[0] -> Option A

4. Quick Check:

   Uninitialized ways[0] can cause issues [OK]

Hint: Always initialize all array elements before use [OK]

Common Mistakes:

• Ignoring ways[0] initialization
• Assuming no undefined behavior without ways[0]
• Starting loop incorrectly

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 modification to the standard climbing stairs approach is correct?

hard

A. Use ways[i] = ways[i-1] + ways[i-3] with base cases ways[0]=1, ways[1]=1, ways[2]=2

B. Use ways[i] = ways[i-1] * ways[i-2] * ways[i-3] with base cases ways[0]=1, ways[1]=1, ways[2]=2

C. Use ways[i] = ways[i-1] + ways[i-2] with base cases ways[0]=1, ways[1]=1

D. Use ways[i] = ways[i-1] + ways[i-2] + ways[i-3] with base cases ways[0]=1, ways[1]=1, ways[2]=2

Solution

1. Step 1: Understand step size options

   Since steps can be 1, 2, or 3, ways to reach step i include ways to reach i-1, i-2, and i-3.

2. Step 2: Formulate recurrence and base cases

   ways[i] = ways[i-1] + ways[i-2] + ways[i-3]; base cases: ways[0]=1 (one way to stand still), ways[1]=1, ways[2]=2.

3. Final Answer:

   Use ways[i] = ways[i-1] + ways[i-2] + ways[i-3] with base cases ways[0]=1, ways[1]=1, ways[2]=2 -> Option D

4. Quick Check:

   Sum previous 3 ways for steps 1,2,3 [OK]

Hint: Add ways for all allowed step sizes [OK]

Common Mistakes:

• Multiplying ways instead of adding
• Ignoring the 3-step option
• Using wrong base cases