Climbing Stairs Problem in DSA Javascript - Execution Trace

Concept Flow - Climbing Stairs Problem

Start at step 0

↓

Can climb 1 or 2 steps

↓

Calculate ways to reach step n

↓

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

↓

Repeat until step n

↓

Return total ways to reach step n

We start from step 0 and calculate ways to reach each step by adding ways to reach the previous one and two steps, repeating until the target step.

Execution Sample

DSA Javascript

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

This code calculates the number of ways to climb n stairs by summing ways to reach the previous two steps.

Execution Table

Step	Operation	dp Array State	Calculation	Result
0	Initialize dp	[1, 1]	Base cases: dp[0]=1, dp[1]=1	N/A
1	Calculate dp[2]	[1, 1, 2]	dp[2] = dp[1] + dp[0] = 1 + 1	2
2	Calculate dp[3]	[1, 1, 2, 3]	dp[3] = dp[2] + dp[1] = 2 + 1	3
3	Calculate dp[4]	[1, 1, 2, 3, 5]	dp[4] = dp[3] + dp[2] = 3 + 2	5
4	Calculate dp[5]	[1, 1, 2, 3, 5, 8]	dp[5] = dp[4] + dp[3] = 5 + 3	8
5	Return result	[1, 1, 2, 3, 5, 8]	Return dp[5]	8

💡 Reached step n=5, calculation complete

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	Final
i	undefined	2	3	4	5	5
dp	[1,1]	[1,1,2]	[1,1,2,3]	[1,1,2,3,5]	[1,1,2,3,5,8]	[1,1,2,3,5,8]

Key Moments - 3 Insights

Why do we start dp array with [1,1] instead of zeros?

Because dp[0] = 1 means there is 1 way to stand at the bottom without climbing, and dp[1] = 1 means only 1 way to reach the first step. This base case is shown in execution_table row 0.

Why do we add dp[i-1] and dp[i-2] to get dp[i]?

Because to reach step i, you can come from step i-1 by climbing 1 step or from step i-2 by climbing 2 steps. This is the core logic shown in calculation column in execution_table rows 1-4.

What happens if n is 1 or 0?

The function returns dp[n] directly from the initialized array [1,1], so for n=0 or n=1, it returns 1 immediately without looping, as shown in execution_table row 0.

Visual Quiz - 3 Questions

Look at the execution_table at Step 3, what is the value of dp array?

A [1, 1, 2, 3, 5]

B [1, 1, 2, 3]

C [1, 1, 2]

D [1, 1, 2, 3, 5, 8]

💡 Hint

Check the 'dp Array State' column at Step 3 in execution_table

At which step does the loop stop according to the execution_table?

A Step 3

B Step 4

C Step 5

D Step 2

💡 Hint

Look at the last calculation step before returning the result in execution_table

If dp[0] was initialized to 0 instead of 1, how would dp[2] change at Step 1?

A dp[2] would be 2

B dp[2] would be 0

C dp[2] would be 1

D dp[2] would be 3

💡 Hint

Refer to the calculation dp[2] = dp[1] + dp[0] in execution_table Step 1

Concept Snapshot

Climbing Stairs Problem:
- dp array stores ways to reach each step
- dp[0] = 1, dp[1] = 1 base cases
- For i >= 2, dp[i] = dp[i-1] + dp[i-2]
- Result is dp[n], total ways to reach step n
- Time complexity: O(n), Space: O(n)

Full Transcript

The Climbing Stairs Problem counts how many ways to reach the top step by climbing 1 or 2 steps at a time. We use a dp array where dp[i] means ways to reach step i. We start with dp[0] = 1 and dp[1] = 1 as base cases. Then for each step from 2 to n, dp[i] is the sum of dp[i-1] and dp[i-2]. This is because you can get to step i from step i-1 or i-2. The code loops from 2 to n, filling dp. Finally, dp[n] is returned as the total ways. The execution table shows dp array growing step by step, with calculations at each step. Key confusions include why dp starts with [1,1], why we add dp[i-1] and dp[i-2], and what happens for small n. The visual quiz tests understanding of dp array states and loop stopping conditions.

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