Climbing Stairs Problem in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the base cases for climbing stairs.

DSA Javascript

function climbStairs(n) {
  if (n <= 1) return 1;
  let dp = new Array(n + 1).fill(0);
  dp[0] = 1;
  dp[1] = [1];
  // rest of the code
}

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

An

B0

C2

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[1] to 0 which means no ways to climb one stair.

Setting dp[1] to n which is incorrect for base case.

dp[1] should be 1 because there is exactly one way to climb one stair.

2fill in blank

medium

Complete the code to fill dp array using the relation for climbing stairs.

DSA Javascript

for (let i = 2; i <= n; i++) {
  dp[i] = dp[i - 1] [1] dp[i - 2];
}

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

A*

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction which gives wrong count.

Using multiplication which is incorrect here.

Number of ways to reach step i is sum of ways to reach i-1 and i-2 steps.

3fill in blank

hard

Fix the error in the return statement to output the total ways to climb n stairs.

DSA Javascript

return dp[1];

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

A[n]

B[n-1]

C[n+1]

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Returning dp[n-1] which is ways to climb one less stair.

Returning dp[0] which is base case for zero stairs.

The answer is stored in dp[n] which represents ways to climb n stairs.

4fill in blank

hard

Fill both blanks to complete the recursive relation for climbing stairs using memoization.

DSA Javascript

function climb(n, memo) {
  if (n <= 1) return 1;
  if (memo[n] !== undefined) return memo[n];
  memo[n] = climb(n [1], memo) [2] climb(n - 2, memo);
  return memo[n];
}

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

A- 1

B+

C- 2

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of addition.

Using wrong subtraction values.

We call climb(n-1) and climb(n-2) and add their results to get ways for n.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each step to its number of ways if ways are more than 1.

DSA Javascript

const waysMap = Object.fromEntries(
  Object.entries(dp).filter(([[1], [2]]) => [2] [3] 1)
);

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

Astep

Bways

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names in destructuring.

Using less than instead of greater than.

We filter entries where ways (value) is greater than 1 and keep step as key.

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