Climbing Stairs Problem in DSA Javascript - Build from Scratch

Climbing Stairs Problem

📖 Scenario: Imagine you are climbing a staircase that has a certain number of steps. You can climb either 1 step or 2 steps at a time. You want to find out how many different ways you can reach the top.

🎯 Goal: Build a program that calculates the number of distinct ways to climb to the top of a staircase with n steps, where you can take either 1 or 2 steps at a time.

📋 What You'll Learn

Create a variable n to represent the total number of steps in the staircase.

Create a variable ways to store the number of ways to climb the stairs.

Use a loop to calculate the number of ways to climb the stairs using the logic that ways to reach step i is the sum of ways to reach steps i-1 and i-2.

Print the total number of ways to climb the staircase.

💡 Why This Matters

🌍 Real World

This problem models situations where you have multiple ways to reach a goal by taking different sized steps, like climbing stairs or moving through stages.

💼 Career

Understanding this problem helps in learning dynamic programming, which is useful in software engineering roles involving optimization and algorithm design.

1

Set the total number of steps

Create a variable called n and set it to 5 to represent the total number of steps in the staircase.

DSA Javascript

# Your code here

Hint

Use const n = 5; to set the number of steps.

2

Initialize the ways array

Create an array called ways with length n + 1. Set ways[0] to 1 and ways[1] to 1 to represent the base cases.

DSA Javascript

const n = 5;
// Your code here

Hint

Initialize the ways array with zeros and set the first two values to 1.

3

Calculate ways to climb

Use a for loop with variable i starting from 2 up to n. Inside the loop, set ways[i] to ways[i - 1] + ways[i - 2].

DSA Javascript

const n = 5;
const ways = new Array(n + 1).fill(0);
ways[0] = 1;
ways[1] = 1;
// Your code here

Hint

Use the formula ways[i] = ways[i - 1] + ways[i - 2] inside the loop.

4

Print the total ways

Print the value of ways[n] using console.log to show the total number of ways to climb the staircase.

DSA Javascript

const n = 5;
const ways = new Array(n + 1).fill(0);
ways[0] = 1;
ways[1] = 1;
for (let i = 2; i <= n; i++) {
    ways[i] = ways[i - 1] + ways[i - 2];
}
// Your code here

Hint

Use console.log(ways[n]); to print the result.

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