Climbing Stairs Problem in DSA Java - Execution Trace

Concept Flow - Climbing Stairs Problem

Start at step 0

↓

Can climb 1 or 2 steps

↓

Calculate ways to reach step i

↓

ways[i

↓

Repeat for all steps up to n

↓

Return ways[n

Start from step 0, for each step calculate ways to reach it by summing ways to reach previous two steps, repeat until step n.

Execution Sample

DSA Java

int climbStairs(int n) {
  int[] ways = new int[n+1];
  ways[0] = 1;
  if (n >= 1) ways[1] = 1;
  for (int i = 2; i <= n; i++) {
    ways[i] = ways[i-1] + ways[i-2];
  }
  return ways[n];
}

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

Execution Table

Step	Operation	Index i	ways[i-2]	ways[i-1]	ways[i]	Visual State (ways array)
1	Initialize base cases	-	-	-	-	[1, 1, _, _, _, _, _]
2	Calculate ways[2]	2	ways[0]=1	ways[1]=1	2	[1, 1, 2, _, _, _, _]
3	Calculate ways[3]	3	ways[1]=1	ways[2]=2	3	[1, 1, 2, 3, _, _, _]
4	Calculate ways[4]	4	ways[2]=2	ways[3]=3	5	[1, 1, 2, 3, 5, _, _]
5	Calculate ways[5]	5	ways[3]=3	ways[4]=5	8	[1, 1, 2, 3, 5, 8, _]
6	Calculate ways[6]	6	ways[4]=5	ways[5]=8	13	[1, 1, 2, 3, 5, 8, 13]
7	Return result ways[6]	-	-	-	13	[1, 1, 2, 3, 5, 8, 13]

💡 Reached step n=6, calculation complete, return ways[6]=13

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	Final
i	-	2	3	4	5	6	6
ways[0]	1	1	1	1	1	1	1
ways[1]	1	1	1	1	1	1	1
ways[2]	_	2	2	2	2	2	2
ways[3]	_	_	3	3	3	3	3
ways[4]	_	_	_	5	5	5	5
ways[5]	_	_	_	_	8	8	8
ways[6]	_	_	_	_	_	13	13

Key Moments - 3 Insights

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

Because ways[0] represents the base case of standing at the bottom (one way to be there), and ways[1] means only one way to reach the first step (one single step). This is shown in execution_table row 1.

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. So total ways to i is sum of ways to i-1 and i-2, as shown in rows 2 to 6.

What happens if n is 1 or 0?

If n=0 or n=1, the function returns 1 immediately because ways[0] and ways[1] are set to 1, meaning only one way to stand or take one step. This is the base case before the loop.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the value of ways[4]?

A 5

B 3

C 8

D 2

💡 Hint

Check the 'ways[i]' column in execution_table row 4.

At which step does the loop stop calculating ways?

A Step 5

B Step 6

C Step 7

D Step 4

💡 Hint

Look at the last calculation step before returning result in execution_table.

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

A ways[2] would be 0

B ways[2] would be 2

C ways[2] would be 1

D ways[2] would be 3

💡 Hint

ways[2] = ways[0] + ways[1], check variable_tracker for ways[0] and ways[1].

Concept Snapshot

Climbing Stairs Problem:
- ways[0] = 1, ways[1] = 1 base cases
- For i from 2 to n: ways[i] = ways[i-1] + ways[i-2]
- Result is ways[n]
- Counts total distinct ways to climb n stairs taking 1 or 2 steps
- Uses bottom-up dynamic programming array

Full Transcript

This visualization shows how to solve the Climbing Stairs Problem using dynamic programming. We start with base cases ways[0] = 1 and ways[1] = 1, meaning one way to stand at the bottom and one way to reach the first step. Then for each step i from 2 to n, we calculate ways[i] by adding ways[i-1] and ways[i-2], because you can reach step i from either step i-1 or i-2. The execution table traces each step's calculation and the ways array state. The variable tracker shows how the index i and ways array values change after each iteration. Key moments clarify why base cases are set and why we add previous two steps. The quiz tests understanding of array values at specific steps and effects of changing base cases. The concept snapshot summarizes the approach as a bottom-up dynamic programming solution counting distinct ways to climb stairs.

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 take at once

B. The minimum number of steps to reach the top

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

D. The number of steps you skip

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 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 C

4. Quick Check:

   Counting ways = total distinct sequences [OK]

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

Common Mistakes:

• Confusing number of ways with minimum steps
• Thinking about maximum steps allowed
• Counting skipped steps instead of ways

2. Which of the following Java method signatures correctly represents a function to solve the Climbing Stairs problem for n steps?

easy

A. public int climbStairs(int n)

B. public void climbStairs(int n)

C. public int climbStairs()

D. public int climbStairs(String n)

Solution

1. Step 1: Check method return type and parameters

   The method should return an integer count and accept an integer input for steps.

2. Step 2: Validate each option

   public int climbStairs(int n) returns int and takes int n, which fits the problem. Others either have wrong return type or parameter type.

3. Final Answer:

   public int climbStairs(int n) -> Option A

4. Quick Check:

   Return int count, input int steps [OK]

Hint: Return int, input int n for steps [OK]

Common Mistakes:

• Using void return type instead of int
• Missing input parameter
• Using wrong parameter type like String

3. What is the output of the following Java code snippet?

public class Solution {
  public static int climbStairs(int n) {
    if (n <= 2) return n;
    int a = 1, b = 2, c = 0;
    for (int i = 3; i <= n; i++) {
      c = a + b;
      a = b;
      b = c;
    }
    return c;
  }
  public static void main(String[] args) {
    System.out.println(climbStairs(4));
  }
}

medium

A. 3

B. 7

C. 4

D. 5

Solution

1. Step 1: Trace the loop for n=4

   Initial a=1, b=2. Loop runs for i=3 and i=4.
   i=3: c=1+2=3, a=2, b=3
   i=4: c=2+3=5, a=3, b=5

2. Step 2: Return value after loop

   Return c=5, which is the number of ways to climb 4 steps.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Ways(4) = 5 [OK]

Hint: Use Fibonacci-like addition for steps > 2 [OK]

Common Mistakes:

• Returning b instead of c
• Off-by-one errors in loop
• Misunderstanding base cases

4. Identify the error in the following Java code for climbing stairs:

public int climbStairs(int n) {
  if (n == 1) return 1;
  int[] dp = new int[n];
  dp[0] = 1;
  dp[1] = 2;
  for (int i = 2; i <= n; i++) {
    dp[i] = dp[i-1] + dp[i-2];
  }
  return dp[n];
}

medium

A. Wrong return statement should be dp[n-1]

B. Array index out of bounds due to dp size and loop range

C. Incorrect base case for n == 1

D. Missing initialization of dp[2]

Solution

1. Step 1: Check array size and indexing

   Array dp size is n, but loop runs i from 2 to n inclusive, accessing dp[n], which is out of bounds.

2. Step 2: Identify correct loop range and return

   Loop should run i < n, and return dp[n-1] to avoid index error.

3. Final Answer:

   Array index out of bounds due to dp size and loop range -> Option B

4. Quick Check:

   Loop index exceeds array size [OK]

Hint: Array size n means max index n-1 [OK]

Common Mistakes:

• Looping to i <= n instead of i < n
• Returning dp[n] instead of dp[n-1]
• Incorrect base case handling

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 approach correctly adapts the Climbing Stairs problem?

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] * dp[i-3] with base dp[0]=1

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

D. Use recursion without memoization for all steps

Solution

1. Step 1: Understand the new step options

   Since you can take 1, 2, or 3 steps, ways to reach step i depend on previous 3 steps.

2. Step 2: Formulate dp relation and base cases

   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 dp[0]=1, dp[1]=1, dp[2]=2 -> Option A

4. Quick Check:

   Include all step sizes in dp sum [OK]

Hint: Sum dp of last 3 steps for 1,2,3 step moves [OK]

Common Mistakes:

• Ignoring 3-step moves in dp formula
• Multiplying dp values instead of adding
• Using recursion without caching causing slow run