Climbing Stairs Problem in DSA Go - Time & Space Complexity

Time Complexity: Climbing Stairs Problem

O(n)

Understanding Time Complexity

We want to understand how the time needed to solve the climbing stairs problem grows as the number of stairs increases.

Specifically, how does the number of steps to calculate ways change when the stairs get longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func climbStairs(n int) int {
    if n <= 2 {
        return n
    }
    dp := make([]int, n+1)
    dp[1], dp[2] = 1, 2
    for i := 3; i <= n; i++ {
        dp[i] = dp[i-1] + dp[i-2]
    }
    return dp[n]
}
    

This code calculates how many distinct ways there are to climb n stairs, taking 1 or 2 steps at a time, using a bottom-up approach.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that fills the dp array from 3 to n.
• How many times: It runs (n - 2) times, once for each stair from 3 up to n.

How Execution Grows With Input

Each new stair requires one addition operation to find the number of ways.

Input Size (n)	Approx. Operations
10	8 additions
100	98 additions
1000	998 additions

Pattern observation: The number of operations grows roughly in a straight line as n increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to compute the number of ways grows directly in proportion to the number of stairs.

Common Mistake

[X] Wrong: "Because the problem uses recursion, it must be exponential time."

[OK] Correct: This code uses a loop with stored results (dynamic programming), so it only calculates each step once, avoiding repeated work.

Interview Connect

Understanding this problem helps you see how to turn slow recursive solutions into fast ones using memory to save work, a key skill in many coding challenges.

Self-Check

"What if we changed the allowed steps from 1 or 2 to 1, 2, or 3 steps? How would the time complexity change?"

Practice

1. What does the Climbing Stairs problem ask you to find?
Given n steps, 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 steps you skip

Solution

1. Step 1: Understand the problem statement

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

2. Step 2: Identify what is being counted

   We count all possible sequences of 1-step and 2-step moves that sum to n steps.

3. Final Answer:

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

4. Quick Check:

   Counting sequences of 1 or 2 steps = total ways [OK]

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

Common Mistakes:

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

2. Which Go function signature correctly represents a function to solve the Climbing Stairs problem for n steps?

easy

A. func climbStairs(n int) bool

B. func climbStairs(n string) string

C. func climbStairs() int

D. func climbStairs(n int) int

Solution

1. Step 1: Identify input and output types

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

2. Step 2: Match function signature

   Only func climbStairs(n int) int matches input int and output int.

3. Final Answer:

   func climbStairs(n int) int -> Option D

4. Quick Check:

   Input int, output int = correct signature [OK]

Hint: Input and output both must be int for steps and ways [OK]

Common Mistakes:

• Using string instead of int for steps
• Missing input parameter
• Returning bool instead of int

3. What is the output of this Go code for n = 4?

func climbStairs(n int) int {
    if n <= 2 {
        return n
    }
    a, b := 1, 2
    for i := 3; i <= n; i++ {
        a, b = b, a+b
    }
    return b
}

fmt.Println(climbStairs(4))

medium

A. 7

B. 6

C. 5

D. 4

Solution

1. Step 1: Trace initial values and loop

   Start with a=1, b=2. Loop i=3 to 4:
   i=3: a=2, b=3
   i=4: a=3, b=5

2. Step 2: Return final value

   Return b=5 after loop ends.

3. Final Answer:

   5 -> Option C

4. Quick Check:

   Ways for 4 steps = 5 [OK]

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

Common Mistakes:

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

4. Find the error in this Go code for climbing stairs:

func climbStairs(n int) int {
    if n == 1 {
        return 1
    }
    a, b := 1, 2
    for i := 2; i <= n; i++ {
        a, b = b, a+b
    }
    return b
}

medium

A. Return statement should return a instead of b

B. Loop starts at wrong index causing extra iteration

C. Initial values of a and b are incorrect

D. Missing base case for n == 2

Solution

1. Step 1: Analyze loop start and iterations

   Loop starts at i=2 but initial values already cover steps 1 and 2, causing one extra iteration.

2. Step 2: Correct loop start

   Loop should start at i=3 to avoid counting extra ways.

3. Final Answer:

   Loop starts at wrong index causing extra iteration -> Option B

4. Quick Check:

   Loop index off by one causes wrong count [OK]

Hint: Start loop at 3 when initial values cover steps 1 and 2 [OK]

Common Mistakes:

• Starting loop at 2 instead of 3
• Changing initial values incorrectly
• Returning wrong variable

5. You want to find the number of ways to climb 10 steps but can take 1, 2, or 3 steps at a time. How would you modify the climbing stairs approach?

hard

A. Use dynamic programming with dp[i] = dp[i-1] + dp[i-2] + dp[i-3]

B. Use the same formula dp[i] = dp[i-1] + dp[i-2]

C. Only count ways taking 3 steps at a time

D. Multiply the number of ways for 1 and 2 steps

Solution

1. Step 1: Understand the new step options

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

2. Step 2: Formulate recurrence relation

   Number of ways dp[i] = dp[i-1] + dp[i-2] + dp[i-3] to include all possibilities.

3. Final Answer:

   Use dynamic programming with dp[i] = dp[i-1] + dp[i-2] + dp[i-3] -> Option A

4. Quick Check:

   Include all step sizes in sum for dp [OK]

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

Common Mistakes:

• Using old formula for only 1 or 2 steps
• Ignoring 3-step moves
• Multiplying counts instead of adding