Climbing Stairs Problem in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Recursive Climbing Stairs Function

What is the output of this Rust code that calculates the number of ways to climb 4 stairs using recursion?

DSA Rust

fn climb_stairs(n: i32) -> i32 {
    if n <= 2 {
        return n;
    }
    climb_stairs(n - 1) + climb_stairs(n - 2)
}

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

A 5

B 7

C 6

D 4

💡 Hint

Think about the base cases and how the function sums ways for n-1 and n-2 steps.

The function returns the number of ways to climb n stairs by summing ways to climb n-1 and n-2 stairs. For n=4, ways are 5.

❓ Predict Output

intermediate

Output of Iterative Climbing Stairs Function

What is the output of this Rust code that calculates the number of ways to climb 5 stairs using iteration?

DSA Rust

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

fn main() {
    println!("{}", climb_stairs(5));
}

A 10

B 7

C 5

D 8

💡 Hint

Iteratively sum the ways to climb previous two steps.

The iterative method sums ways for previous two steps. For n=5, the output is 8.

🔧 Debug

advanced

Identify the Runtime Error in Climbing Stairs Code

What error does this Rust code produce when run, and why?

DSA Rust

fn climb_stairs(n: i32) -> i32 {
    if n == 0 {
        return 0;
    }
    if n == 1 {
        return 1;
    }
    climb_stairs(n - 1) + climb_stairs(n - 2)
}

fn main() {
    println!("{}", climb_stairs(-1));
}

A No error, outputs 0

B Compile-time error: mismatched types

C Stack overflow due to infinite recursion

D Runtime panic: index out of bounds

💡 Hint

Consider what happens when n is negative.

The function calls itself with decreasing n without base case for negative values, causing infinite recursion and stack overflow.

🧠 Conceptual

advanced

Time Complexity of Climbing Stairs Recursive Solution

What is the time complexity of the naive recursive solution for the climbing stairs problem with input n?

A O(n)

B O(2^n)

C O(n^2)

D O(log n)

💡 Hint

Think about how many calls the recursion makes for each step.

The naive recursion calls itself twice for each n, leading to exponential growth in calls.

🚀 Application

expert

Number of Ways to Climb Stairs with Variable Steps

Given the Rust function below that calculates ways to climb n stairs with steps of size 1, 3, or 5, what is the output for n = 6?

DSA Rust

fn climb_stairs_var(n: usize) -> usize {
    let mut dp = vec![0; n + 1];
    dp[0] = 1;
    for i in 1..=n {
        if i >= 1 {
            dp[i] += dp[i - 1];
        }
        if i >= 3 {
            dp[i] += dp[i - 3];
        }
        if i >= 5 {
            dp[i] += dp[i - 5];
        }
    }
    dp[n]
}

fn main() {
    println!("{}", climb_stairs_var(6));
}

A 8

B 12

C 15

D 10

💡 Hint

Use dynamic programming to sum ways from steps 1, 3, and 5 behind current step.

The dp array counts ways to reach each step by summing ways from steps 1, 3, and 5 behind. For n=6, total ways are 8.

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