Why Climbing Stairs Problem in DSA Typescript?

The Big Idea

What if you could find all ways to climb stairs without trying every step combination?

The Scenario

Imagine you want to climb a staircase with many steps, but you can only take one or two steps at a time. You want to know how many different ways you can reach the top.

Without a smart method, you might try to list all possible step combinations by hand.

The Problem

Listing all step combinations manually is slow and confusing, especially as the number of steps grows. You might forget some combinations or repeat others, making it error-prone and frustrating.

The Solution

The Climbing Stairs Problem uses a simple counting method that builds on smaller results to find the total ways quickly. It avoids listing all combinations by using addition of previous counts, making it fast and reliable.

Before vs After

✗ Before

function countWays(steps: number): number {
  // Manually list all combinations (impractical)
  // For example, for 3 steps: (1,1,1), (1,2), (2,1)
  return 3; // Hardcoded for small steps
}

✓ After

function countWays(steps: number): number {
  if (steps <= 2) return steps;
  let waysOneStepBefore = 2;
  let waysTwoStepsBefore = 1;
  let totalWays = 0;
  for (let i = 3; i <= steps; i++) {
    totalWays = waysOneStepBefore + waysTwoStepsBefore;
    waysTwoStepsBefore = waysOneStepBefore;
    waysOneStepBefore = totalWays;
  }
  return totalWays;
}

What It Enables

This method lets you quickly find how many ways to climb any number of steps without listing all possibilities.

Real Life Example

Planning how many ways to reach the top of a staircase when you can hop one or two steps at a time, like in a fitness app counting exercise variations.

Key Takeaways

Manual counting is slow and error-prone for many steps.

Using previous results speeds up counting ways to climb.

This approach scales easily to large numbers of steps.

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 maximum number of steps you can take at once

B. The minimum number of steps to reach the top

C. The number of stairs you skip

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

Solution

1. Step 1: Understand the problem statement

   The problem asks for the count of different ways to climb n stairs taking 1 or 2 steps at a time.

2. Step 2: Identify what is being asked

   We want the total number of distinct sequences of 1-step and 2-step moves that sum to n.

3. Final Answer:

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

4. Quick Check:

   Count ways = total distinct sequences [OK]

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

Common Mistakes:

• Confusing number of ways with minimum steps
• Thinking about maximum steps allowed
• Counting skipped stairs instead of ways

2. Which of the following TypeScript function signatures correctly represents a function to solve the Climbing Stairs problem?

easy

A. function climbStairs(n: number): number

B. function climbStairs(n: string): string

C. function climbStairs(): void

D. function climbStairs(n: number): void

Solution

1. Step 1: Identify input and output types

   The input is the number of stairs (number), output is the count of ways (number).

2. Step 2: Match function signature

   Only the signature with input number and output number matches the problem requirements.

3. Final Answer:

   function climbStairs(n: number): number -> Option A

4. Quick Check:

   Input and output types match problem [OK]

Hint: Input and output both must be numbers [OK]

Common Mistakes:

• Using string type for number input
• Function returning void instead of number
• No input parameter in function

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

function climbStairs(n: number): number {
  if (n <= 2) return n;
  let a = 1, b = 2;
  for (let i = 3; i <= n; i++) {
    let 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: Trace initial values and loop

   For n=4, start with a=1, b=2. Loop i=3 to 4: - i=3: temp=1+2=3, a=2, b=3 - i=4: temp=2+3=5, a=3, b=5

2. Step 2: Return final value

   Return b=5 as the number of ways to climb 4 stairs.

3. Final Answer:

   5 -> Option B

4. Quick Check:

   Ways for 4 stairs = 5 [OK]

Hint: Use Fibonacci-like addition for steps count [OK]

Common Mistakes:

• Returning a instead of b
• Off-by-one error in loop
• Incorrect initial values for a and b

4. Identify the error in this TypeScript code for climbing stairs:

function climbStairs(n: number): number {
  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. Initial values of a and b are incorrect

B. Return statement should return a instead of b

C. Loop starts at wrong index causing off-by-one error

D. Missing base case for n == 2

Solution

1. Step 1: Analyze loop start and base cases

   Loop starts at i=2 but initial values assume i=3 start. This causes extra iteration and wrong count.

2. Step 2: Correct loop start

   Loop should start at i=3 to align with initial values a=1, b=2 and base case for n=2.

3. Final Answer:

   Loop starts at wrong index causing off-by-one error -> Option C

4. Quick Check:

   Loop index must match initial conditions [OK]

Hint: Check loop start matches base cases [OK]

Common Mistakes:

• Starting loop at 2 instead of 3
• Ignoring base case for n=2
• Swapping a and b updates

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

hard

A. Use dp[i] = dp[i-1] + dp[i-2] + dp[i-3] with base dp[0]=1, dp[1]=1, dp[2]=2

B. Use dp[i] = dp[i-1] * dp[i-2] * dp[i-3] with base dp[0]=1, dp[1]=1, dp[2]=2

C. Use dp[i] = dp[i-1] + dp[i-2] only with base dp[0]=1, dp[1]=1

D. Use dp[i] = dp[i-1] + dp[i-3] with base dp[0]=1, dp[1]=1, dp[2]=2

Solution

1. Step 1: Understand the new step options

   Now steps can be 1, 2, or 3 at a time, so ways to reach i depend on previous 3 states.

2. Step 2: Formulate recurrence relation

   Number of ways to reach step i = sum of ways to reach i-1, i-2, and i-3.

3. Step 3: Set base cases

   dp[0]=1 (one way to stand still), dp[1]=1, dp[2]=2 (1+1 or 2 steps).

4. Final Answer:

   Use dp[i] = dp[i-1] + dp[i-2] + dp[i-3] with base dp[0]=1, dp[1]=1, dp[2]=2 -> Option A

5. Quick Check:

   Sum previous 3 states for new step sizes [OK]

Hint: Sum ways from last 3 steps for 1,2,3 moves [OK]

Common Mistakes:

• Multiplying previous states instead of adding
• Ignoring the 3-step option
• Wrong base cases for dp array