Climbing Stairs Problem in DSA Python - 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 changes as the number of stairs grows.

Specifically, how many steps does the program take to find the total ways to climb?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def climb_stairs(n):
    if n <= 2:
        return n
    dp = [0] * (n + 1)
    dp[1], dp[2] = 1, 2
    for i in range(3, n + 1):
        dp[i] = dp[i - 1] + dp[i - 2]
    return dp[n]

This code calculates the number of ways 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 to n.

How Execution Grows With Input

As the number of stairs increases, the loop runs more times, adding one operation per stair.

Input Size (n)	Approx. Operations
10	8 (loop runs from 3 to 10)
100	98
1000	998

Pattern observation: The operations grow roughly in a straight line as n increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the number of ways grows directly with the number of stairs.

Common Mistake

[X] Wrong: "Because the problem looks like Fibonacci, it must take exponential time."

[OK] Correct: This code uses a loop and stores results, so it avoids repeating work and runs quickly in linear time.

Interview Connect

Understanding this problem’s time complexity shows you can spot when to use dynamic programming to improve efficiency, a key skill in many coding challenges.

Self-Check

"What if we changed the code to use simple recursion without storing results? 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 steps climbed

B. The maximum number of steps you can climb at once

C. The minimum number of steps to reach the top

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

4. Quick Check:

   Count ways to climb = number of 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 instead of ways
• Counting total steps instead of sequences

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

easy

A. def climb_stairs(n: int) -> int:

B. def climb_stairs(n): return str

C. def climb_stairs(n: str) -> int:

D. def climb_stairs() -> int:

Solution

1. Step 1: Check parameter and return types

   The input n is an integer representing steps, and output is an integer count of ways.

2. Step 2: Match function signature

   def climb_stairs(n: int) -> int: correctly uses int input and int output with type hints.

3. Final Answer:

   def climb_stairs(n: int) -> int: -> Option A

4. Quick Check:

   Input int, output int function signature [OK]

Hint: Input is int steps, output is int ways [OK]

Common Mistakes:

• Using string type for input
• Returning string instead of int
• Missing input parameter

3. What is the output of this code?

def climb_stairs(n: int) -> int:
    if n <= 2:
        return n
    a, b = 1, 2
    for _ in range(3, n + 1):
        a, b = b, a + b
    return b

print(climb_stairs(5))

medium

A. 5

B. 8

C. 13

D. 10

Solution

1. Step 1: Trace initial values and base cases

   For n=5, start with a=1, b=2. Loop runs for i=3 to 5.

2. Step 2: Calculate loop iterations

   Iteration 3: a=2, b=3; Iteration 4: a=3, b=5; Iteration 5: a=5, b=8.

3. Final Answer:

   8 -> Option B

4. Quick Check:

   Ways to climb 5 steps = 8 [OK]

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

Common Mistakes:

• Returning n for all inputs
• Off-by-one errors in loop
• Mixing a and b values

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

def climb_stairs(n: int) -> int:
    if n == 1:
        return 1
    a, b = 1, 2
    for i in range(2, n):
        a = b
        b = a + b
    return b

print(climb_stairs(4))

medium

A. Variables a and b are updated incorrectly

B. Loop range should start from 3

C. Base case for n == 2 is missing

D. Return statement should be 'return a'

Solution

1. Step 1: Analyze variable updates inside loop

   Inside loop, a is set to b, then b = a + b uses updated a, causing wrong sum.

2. Step 2: Correct update order

   We must update both simultaneously like a, b = b, a + b to avoid using updated a prematurely.

3. Final Answer:

   Variables a and b are updated incorrectly -> Option A

4. Quick Check:

   Simultaneous update needed for correct Fibonacci logic [OK]

Hint: Update a,b together to avoid wrong sums [OK]

Common Mistakes:

• Updating variables sequentially causing wrong values
• Wrong loop range causing missed steps
• Missing base cases

5. You want to find the number of ways to climb 10 steps but can take 1, 2, or 3 steps at a time. Which modification to the standard climbing stairs solution 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] with base dp[0]=1, dp[1]=1

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

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 step options and recurrence

   With steps 1, 2, or 3 allowed, ways to reach i = sum of ways to reach i-1, i-2, and i-3.

2. Step 2: Set correct base cases

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

3. 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 C

4. Quick Check:

   Sum previous three dp values for 3-step climb [OK]

Hint: Sum ways from last 3 steps for 1-2-3 step climb [OK]

Common Mistakes:

• Multiplying dp values instead of adding
• Using only two previous dp values
• Incorrect base cases