Climbing Stairs Problem in DSA Rust - 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 does the number of steps to calculate ways to climb increase with more stairs?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn climb_stairs(n: i32) -> i32 {
    if n <= 2 {
        return n;
    }
    let mut first = 1;
    let mut second = 2;
    for _ in 3..=n {
        let third = first + second;
        first = second;
        second = third;
    }
    second
}

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that runs from 3 to n, adding two numbers each time.
• How many times: It runs (n - 2) times, once for each stair from 3 up to n.

How Execution Grows With Input

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

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 find the number of ways grows directly in proportion to the number of stairs.

Common Mistake

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

[OK] Correct: This code uses a simple loop and stores only two previous results, so it runs quickly in a straight line, not exploding exponentially.

Interview Connect

Understanding this problem's time complexity shows you can spot when a recursive idea can be turned into a fast loop, a key skill in coding interviews.

Self-Check

"What if we changed the problem to allow steps of 1, 2, or 3 at a time? 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 maximum number of steps you can skip

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

C. The minimum number of steps to reach the top

D. The number of ways to climb only 1 step at a time

Solution

1. Step 1: Understand the problem statement

   The problem asks for the count of different ways to reach the top by taking either 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 B

4. Quick Check:

   Counting distinct sequences = total ways [OK]

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

Common Mistakes:

• Confusing number of ways with minimum steps
• Counting only 1-step moves
• Ignoring 2-step moves

2. Which of the following Rust function signatures correctly represents a function to solve the Climbing Stairs problem for input n: i32 returning an i32?

easy

A. fn climb_stairs(n: i32) -> String

B. fn climb_stairs(n: i32, result: i32)

C. fn climb_stairs(n: String) -> i32

D. fn climb_stairs(n: i32) -> i32

Solution

1. Step 1: Check parameter and return types

   The input is number of steps as i32, and output is number of ways as i32.

2. Step 2: Match function signature

   Only fn climb_stairs(n: i32) -> i32 matches input and output types correctly.

3. Final Answer:

   fn climb_stairs(n: i32) -> i32 -> Option D

4. Quick Check:

   Input and output types match function signature [OK]

Hint: Input and output types must match problem data [OK]

Common Mistakes:

• Using wrong return type like String
• Passing input as String instead of i32
• Missing return type or extra parameters

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

fn climb_stairs(n: i32) -> i32 {
    if n <= 2 {
        return n;
    }
    let mut first = 1;
    let mut second = 2;
    let mut ways = 0;
    for _ in 3..=n {
        ways = first + second;
        first = second;
        second = ways;
    }
    ways
}

fn main() {
    println!("{}", climb_stairs(4));
}

medium

A. 4

B. 6

C. 5

D. 7

Solution

1. Step 1: Trace initial values and loop

   n=4, first=1, second=2. Loop runs for i=3 and i=4. - i=3: ways=1+2=3, first=2, second=3 - i=4: ways=2+3=5, first=3, second=5

2. Step 2: Return final ways value

   After loop, ways=5 is returned and printed.

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 wrong initial value
• Off-by-one errors in loop
• Mixing first and second variables

4. Identify the bug in this Rust code for climbing stairs and select the fix:

fn climb_stairs(n: i32) -> i32 {
    if n == 1 {
        return 1;
    }
    let mut first = 1;
    let mut second = 2;
    let mut ways = 0;
    for i in 2..n {
        ways = first + second;
        first = second;
        second = ways;
    }
    ways
}

medium

A. Return second instead of ways at the end

B. Initialize ways to 1 instead of 0

C. Change if condition to n <= 2

D. Change loop to for i in 3..=n

Solution

1. Step 1: Identify the issue with base case n=2

   For n=2, if not triggered, loop 2..2 empty, returns ways=0 (wrong, should be 2). second initialized to 2.

2. Step 2: Observe the update logic

   Each loop sets second to the new ways value (nth ways). After loop (or none), second is correct: initial 2 for n=2, updated for larger n.

3. Final Answer:

   Return second instead of ways at the end -> Option A

4. Quick Check:

   second holds nth ways after shifts [OK]

Hint: second tracks the current (nth) ways value [OK]

Common Mistakes:

• Loop runs too short or too long
• Wrong initial if condition
• Returning wrong variable

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

hard

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

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

C. Use dp[i] = dp[i-1] + dp[i-2] only, ignore 3 steps

D. Use recursion without memoization for all steps

Solution

1. Step 1: Understand the new step options

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

2. Step 2: Update recurrence relation and base cases

   Use dp[i] = dp[i-1] + dp[i-2] + dp[i-3]. Base cases: dp[0]=1 (one way to stand still), dp[1]=1, dp[2]=2.

3. Final Answer:

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

4. Quick Check:

   Sum of last 3 dp values for each step [OK]

Hint: Sum ways from last 3 steps for new step sizes [OK]

Common Mistakes:

• Multiplying dp values instead of adding
• Ignoring 3-step moves
• Using recursion without memoization causing slowness