Climbing Stairs Problem in DSA Javascript - 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 when n = 1?

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

intermediate

How do you express the number of ways to climb n stairs using previous results?

Ways(n) = Ways(n-1) + Ways(n-2). You can reach step n by taking 1 step from n-1 or 2 steps from n-2.

intermediate

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

Because the number of ways to climb n stairs follows the same addition pattern as Fibonacci numbers, starting with 1 and 1.

intermediate

What is a simple JavaScript approach to solve the Climbing Stairs Problem efficiently?

Use a loop to build up the number of ways from the base cases, storing only the last two results to save space.

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

A 3

B 4

C 2

D 5

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

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-2)

You can reach step n from step n-1 or n-2, so add those ways.

What are the base cases for the Climbing Stairs Problem?

A Ways(0) = 0, Ways(1) = 1

B Ways(1) = 2, Ways(2) = 3

C Ways(1) = 1, Ways(2) = 2

D Ways(0) = 1, Ways(1) = 1

Ways(0) = 1 (one way to stand at the bottom without climbing), Ways(1) = 1 (one step).

Which data structure is most useful to store intermediate results in this problem?

A Queue

B Array

C Stack

D Tree

An array stores ways for each step efficiently.

What is the time complexity of the iterative solution to the Climbing Stairs Problem?

A O(n)

B O(n^2)

C O(2^n)

D O(log n)

The iterative solution calculates ways for each step once, so linear time.

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

Think about how you can reach step n from previous steps.

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

Look at how the number of ways to climb grows with n.

Practice

1. What does the Climbing Stairs problem ask you to find?
Given n stairs, how many distinct ways can you climb to the top if you can take 1 or 2 steps at a time?

easy

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

B. The minimum number of steps to reach the top

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

D. The number of stairs you skip

Solution

1. Step 1: Understand the problem statement

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

2. Step 2: Identify what is being counted

   We count all possible sequences of 1-step and 2-step moves that sum to n stairs.

3. Final Answer:

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

4. Quick Check:

   Counting sequences = total ways [OK]

Hint: Count sequences of 1 or 2 steps summing to n [OK]

Common Mistakes:

• Thinking it's about minimum steps
• Confusing number of ways with step size
• Ignoring the 2-step option

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

easy

A. function climbStairs(n, m) {}

B. function climb_stairs(n) {}

C. function climbStairs(n) {}

D. function climbStairs() {}

Solution

1. Step 1: Identify the function name and parameters

   The problem requires one input: the number of stairs n. The function name should be descriptive and camelCase is standard in JavaScript.

2. Step 2: Check each option

   function climbStairs(n) {} uses camelCase and one parameter n, which is correct. function climb_stairs(n) {} uses snake_case which is uncommon in JS. function climbStairs(n, m) {} has an extra parameter m which is unnecessary. function climbStairs() {} has no parameters, so it cannot accept n.

3. Final Answer:

   function climbStairs(n) {} -> Option C

4. Quick Check:

   Correct function name and parameter [OK]

Hint: Use camelCase and one parameter n for stairs [OK]

Common Mistakes:

• Using snake_case instead of camelCase
• Adding unnecessary parameters
• Omitting the parameter for stairs

3. What is the output of the following JavaScript code?

function climbStairs(n) {
  if (n <= 2) return n;
  let a = 1, b = 2, temp;
  for (let i = 3; i <= n; i++) {
    temp = a + b;
    a = b;
    b = temp;
  }
  return b;
}
console.log(climbStairs(4));

medium

A. 3

B. 5

C. 4

D. 6

Solution

1. Step 1: Understand the base cases and loop

   For n=1, returns 1; for n=2, returns 2. For n=4, loop runs from i=3 to 4.

2. Step 2: Trace the loop iterations

   i=3: temp = 1+2=3, a=2, b=3
   i=4: temp = 2+3=5, a=3, b=5
   Return b=5.

3. Final Answer:

   5 -> Option B

4. Quick Check:

   Ways for 4 stairs = 5 [OK]

Hint: Trace loop updates for i=3 and i=4 [OK]

Common Mistakes:

• Returning a instead of b
• Miscounting loop iterations
• Confusing base case returns

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

function climbStairs(n) {
  if (n == 1) return 1;
  let a = 1, b = 2;
  for (let i = 2; i <= n; i++) {
    let temp = a + b;
    a = b;
    b = temp;
  }
  return b;
}

medium

A. Loop starts at wrong index causing extra iteration

B. Base case for n=1 is incorrect

C. Variables a and b are initialized incorrectly

D. Return statement should return a instead of b

Solution

1. Step 1: Analyze the loop range

   The loop runs from i=2 to n inclusive. For n=2, loop runs once, but base cases expect no loop for n=2.

2. Step 2: Identify correct loop start

   Loop should start at i=3 to avoid extra iteration and match base cases. Starting at i=2 causes one extra iteration.

3. Final Answer:

   Loop starts at wrong index causing extra iteration -> Option A

4. Quick Check:

   Loop index must start at 3 [OK]

Hint: Loop should start at 3, not 2 [OK]

Common Mistakes:

• Starting loop at 2 instead of 3
• Misunderstanding base cases
• Returning wrong variable

5. You want to find the number of ways to climb 10 stairs but can take 1, 2, or 3 steps at a time. Which modification to the climbing stairs code is correct?

function climbStairs(n) {
  if (n <= 1) return 1;
  let dp = [1, 1, 2];
  for (let i = 3; i <= n; i++) {
    dp[i] = dp[i-1] + dp[i-2] + dp[i-3];
  }
  return dp[n];
}

hard

A. dp array initialization is missing zeros

B. Base cases are incorrect for 3-step option

C. Loop should start at i=2 instead of i=3

D. This code correctly counts ways with 1, 2, or 3 steps

Solution

1. Step 1: Understand the problem extension

   Now steps can be 1, 2, or 3. The number of ways to reach step i is sum of ways to reach i-1, i-2, and i-3.

2. Step 2: Check code logic and base cases

   dp array starts with dp[0]=1 (1 way to stand still), dp[1]=1, dp[2]=2 (ways to climb 2 steps). Loop starts at i=3 and sums previous three dp values correctly.

3. Final Answer:

   This code correctly counts ways with 1, 2, or 3 steps -> Option D

4. Quick Check:

   Sum of last 3 dp values for each step [OK]

Hint: Sum last 3 dp values for each step [OK]

Common Mistakes:

• Not updating base cases for 3-step
• Starting loop at wrong index
• Forgetting to sum three previous dp values