Climbing Stairs Problem in DSA Rust - Execution Trace

Concept Flow - Climbing Stairs Problem

Start at step 0

↓

Calculate ways to step 1

↓

For each step i from 2 to n

↓

ways[i

↓

Repeat until step n

↓

Return ways[n

We start from step 0 and calculate ways to reach each step by adding ways to previous two steps until we reach step n.

Execution Sample

DSA Rust

fn climb_stairs(n: usize) -> usize {
    let mut ways = vec![0; n + 1];
    ways[0] = 1;
    if n > 0 {
        ways[1] = 1;
    }
    for i in 2..=n {
        ways[i] = ways[i - 1] + ways[i - 2];
    }
    ways[n]
}

Calculates number of ways to climb n stairs by summing ways to previous two steps.

Execution Table

Step	Operation	ways Array State	Calculation	Visual State
0	Initialize ways[0]	[1, 0, 0, 0, 0, 0]	ways[0] = 1	1, 0, 0, 0, 0, 0
1	Initialize ways[1]	[1, 1, 0, 0, 0, 0]	ways[1] = 1	1, 1, 0, 0, 0, 0
2	Calculate ways[2]	[1, 1, 2, 0, 0, 0]	ways[2] = ways[1] + ways[0] = 1 + 1	1, 1, 2, 0, 0, 0
3	Calculate ways[3]	[1, 1, 2, 3, 0, 0]	ways[3] = ways[2] + ways[1] = 2 + 1	1, 1, 2, 3, 0, 0
4	Calculate ways[4]	[1, 1, 2, 3, 5, 0]	ways[4] = ways[3] + ways[2] = 3 + 2	1, 1, 2, 3, 5, 0
5	Calculate ways[5]	[1, 1, 2, 3, 5, 8]	ways[5] = ways[4] + ways[3] = 5 + 3	1, 1, 2, 3, 5, 8
6	Return ways[5]	[1, 1, 2, 3, 5, 8]	Result = 8	Final ways array: 1, 1, 2, 3, 5, 8

💡 Reached step n=5, calculation complete, returned ways[5] = 8

Variable Tracker

Variable	Start	After Step 0	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	Final
i	-	0	1	2	3	4	5	5
ways	[0,0,0,0,0,0]	[1,0,0,0,0,0]	[1,1,0,0,0,0]	[1,1,2,0,0,0]	[1,1,2,3,0,0]	[1,1,2,3,5,0]	[1,1,2,3,5,8]	[1,1,2,3,5,8]

Key Moments - 3 Insights

Why do we set ways[0] = 1 at the start?

ways[0] = 1 means there is exactly one way to stand at the bottom without climbing any steps. This base case is needed to build the rest of the solution as shown in execution_table row 0.

Why do we add ways[i-1] and ways[i-2] to get ways[i]?

Because to reach step i, you can come from step i-1 by taking one step or from step i-2 by taking two steps. This is shown in execution_table rows 2 to 5 where ways[i] sums the previous two values.

What happens if n is 1?

If n=1, the loop does not run and ways[1] = 1 is returned directly, meaning only one way to climb one step, as initialized in execution_table row 1.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the value of ways[3]?

A 3

B 2

C 5

D 1

💡 Hint

Check the 'ways Array State' column at step 3 in the execution_table.

At which step does the loop stop running?

A Step 4

B Step 5

C Step 6

D Step 3

💡 Hint

Look at the 'Step' column and note when the return happens in the execution_table.

If ways[0] was set to 0 instead of 1, how would ways[2] change?

A ways[2] would be 2

B ways[2] would be 0

C ways[2] would be 1

D ways[2] would be 3

💡 Hint

Refer to the calculation for ways[2] in execution_table row 2.

Concept Snapshot

Climbing Stairs Problem:
- ways[0] = 1, ways[1] = 1 base cases
- For i >= 2: ways[i] = ways[i-1] + ways[i-2]
- Represents ways to reach step i by 1 or 2 steps
- Result is ways[n], total ways to climb n stairs
- Uses dynamic programming with O(n) time and space

Full Transcript

The Climbing Stairs Problem counts how many ways to reach the top of n stairs by taking 1 or 2 steps at a time. We start by setting ways[0] and ways[1] to 1, meaning one way to stand still and one way to climb the first step. Then for each step from 2 to n, we calculate ways[i] by adding ways[i-1] and ways[i-2], because you can get to step i from either step i-1 or i-2. This builds up the total ways step by step. Finally, we return ways[n] as the answer. The execution table shows each step's calculation and the ways array state. Key moments clarify why base cases are set and how the sum works. The quiz tests understanding of these steps and the array values.

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