Climbing Stairs Problem in DSA Rust

Mental Model

Count how many ways you can reach the top by taking 1 or 2 steps at a time.

Analogy: Imagine climbing stairs where you can take either one step or two steps forward each time. You want to know how many different paths you can take to reach the top.

Start: 0
Steps: 1  2  3  4  5  ...  n
You can jump 1 or 2 steps forward each time.

Dry Run Walkthrough

Input: stairs: 4 steps

Goal: Find how many distinct ways to reach step 4 by taking 1 or 2 steps at a time

Step 1: At step 0, ways to reach = 1 (starting point)

ways[0] = 1

Why: We start at the bottom, so there is exactly one way to be there

Step 2: Calculate ways to reach step 1: ways[1] = ways[0] = 1

ways = [1, 1, _, _, _]

Why: From step 0, only one way to step 1 by taking one step

Step 3: Calculate ways to reach step 2: ways[2] = ways[1] + ways[0] = 1 + 1 = 2

ways = [1, 1, 2, _, _]

Why: Step 2 can be reached from step 1 or step 0

Step 4: Calculate ways to reach step 3: ways[3] = ways[2] + ways[1] = 2 + 1 = 3

ways = [1, 1, 2, 3, _]

Why: Step 3 can be reached from step 2 or step 1

Step 5: Calculate ways to reach step 4: ways[4] = ways[3] + ways[2] = 3 + 2 = 5

ways = [1, 1, 2, 3, 5]

Why: Step 4 can be reached from step 3 or step 2

Result:

ways = [1, 1, 2, 3, 5]
Answer: 5 ways to climb 4 steps

Annotated Code

DSA Rust

fn climb_stairs(n: usize) -> usize {
    if n == 0 || n == 1 {
        return 1;
    }
    let mut ways = vec![0; n + 1];
    ways[0] = 1; // one way to stand at the bottom
    ways[1] = 1; // one way to reach first step
    for i in 2..=n {
        ways[i] = ways[i - 1] + ways[i - 2]; // sum ways from previous two steps
    }
    ways[n]
}

fn main() {
    let stairs = 4;
    let result = climb_stairs(stairs);
    println!("{}", result);
}

if n == 0 || n == 1 { return 1; }

handle base cases where only one way exists

ways[0] = 1; // one way to stand at the bottom

initialize ways to stand at step 0

ways[1] = 1; // one way to reach first step

initialize ways to reach step 1

for i in 2..=n { ways[i] = ways[i - 1] + ways[i - 2]; }

calculate ways to reach step i by summing ways to reach previous two steps

ways[n]

return total ways to reach step n

Output

5

Complexity Analysis

Time: O(n) because we calculate ways for each step once

Space: O(n) because we store ways for all steps up to n

vs Alternative: Compared to recursive approach without memoization which is exponential, this dynamic programming approach is efficient and fast

Edge Cases

n = 0 (no steps)

Returns 1 because standing still counts as one way

DSA Rust

if n == 0 || n == 1 { return 1; }

n = 1 (one step)

Returns 1 because only one way to take one step

DSA Rust

if n == 0 || n == 1 { return 1; }

When to Use This Pattern

When you see a problem asking for number of ways to reach a point with fixed step sizes, use dynamic programming to build up solutions from smaller steps.

Common Mistakes

Mistake: Forgetting to handle base cases n=0 or n=1
Fix: Add explicit checks returning 1 for these base cases

Mistake: Using recursion without memoization causing slow exponential time
Fix: Use bottom-up dynamic programming with an array to store intermediate results

Summary

Counts the number of ways to climb n stairs taking 1 or 2 steps at a time.

Use when you want to find total distinct paths with fixed step increments.

The number of ways to reach step n is the sum of ways to reach steps n-1 and n-2.

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