Climbing Stairs Problem in DSA Javascript

Mental Model

You want to find how many ways you can reach the top of stairs by taking 1 or 2 steps at a time.

Analogy: Imagine climbing a staircase where you can take either one step or two steps at once. Counting all possible ways to reach the top is like counting all different paths you can take.

Start: 0
Steps: 1 2 3 4 5 ... n
You can jump -> 1 step or -> 2 steps at a time

Dry Run Walkthrough

Input: stairs: 4 steps

Goal: Find how many distinct ways to reach step 4 by taking 1 or 2 steps at a time

Step 1: At step 1, ways = 1 (only one way: single step)

ways[1] = 1

Why: Base case: only one way to reach first step

Step 2: At step 2, ways = ways[1] + ways[0] = 1 + 1 = 2

ways[2] = 2

Why: You can reach step 2 from step 1 or step 0

Step 3: At step 3, ways = ways[2] + ways[1] = 2 + 1 = 3

ways[3] = 3

Why: You can reach step 3 from step 2 or step 1

Step 4: At step 4, ways = ways[3] + ways[2] = 3 + 2 = 5

ways[4] = 5

Why: You can reach step 4 from step 3 or step 2

Result:

ways array: [1, 1, 2, 3, 5]
Answer: 5 ways to climb 4 steps

Annotated Code

DSA Javascript

class ClimbingStairs {
  climbStairs(n) {
    if (n <= 1) return 1; // base case: 1 way to climb 0 or 1 step
    const ways = new Array(n + 1).fill(0);
    ways[0] = 1; // 1 way to stand at the bottom
    ways[1] = 1; // 1 way to reach first step
    for (let i = 2; i <= n; i++) {
      ways[i] = ways[i - 1] + ways[i - 2]; // sum ways from previous two steps
    }
    return ways[n];
  }
}

const stairs = new ClimbingStairs();
const n = 4;
console.log(stairs.climbStairs(n));

if (n <= 1) return 1; // base case: 1 way to climb 0 or 1 step

handle smallest stairs cases directly

ways[0] = 1; // 1 way to stand at the bottom

initialize ways to stand still before climbing

ways[1] = 1; // 1 way to reach first step

initialize ways to reach first step

ways[i] = ways[i - 1] + ways[i - 2]; // sum ways from previous two steps

calculate ways to current step by adding ways from last two steps

Output

5

Complexity Analysis

Time: O(n) because we calculate ways for each step once

Space: O(n) because we store ways for all steps up to n

vs Alternative: Compared to recursive approach without memoization which is exponential, this dynamic programming approach is efficient and fast

Edge Cases

n = 0 (no steps)

Returns 1 because standing still counts as one way

DSA Javascript

if (n <= 1) return 1; // base case: 1 way to climb 0 or 1 step

n = 1 (one step)

Returns 1 because only one way to climb one step

DSA Javascript

if (n <= 1) return 1; // base case: 1 way to climb 0 or 1 step

When to Use This Pattern

When you see a problem asking for number of ways to reach a point with fixed step sizes, use dynamic programming to build solutions from smaller steps.

Common Mistakes

Mistake: Not initializing base cases ways[0] and ways[1], causing wrong results
Fix: Always set ways[0] = 1 and ways[1] = 1 before loop

Mistake: Using recursion without memoization leading to slow exponential time
Fix: Use bottom-up dynamic programming with an array to store intermediate results

Summary

Calculates how many distinct ways to climb n stairs taking 1 or 2 steps at a time.

Use when counting paths with fixed step increments or similar combinatorial problems.

The number of ways to reach step n equals the sum of ways to reach steps n-1 and n-2.

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