Climbing Stairs Problem in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the base case for 1 step.

DSA C

if (n == 1) return [1];

Drag options to blanks, or click blank then click option'

A2

B0

Cn

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 instead of 1 for the base case.

Returning n which is unnecessary here.

When there is only 1 step, there is exactly 1 way to climb it.

2fill in blank

medium

Complete the code to return the base case for 2 steps.

DSA C

if (n == 2) return [1];

Drag options to blanks, or click blank then click option'

A1

B3

C2

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 1 instead of 2 for the base case.

Returning n which is 2 but not explained.

For 2 steps, you can climb as (1+1) or (2), so 2 ways.

3fill in blank

hard

Fix the error in the recursive return statement to correctly sum the ways.

DSA C

return climbStairs(n [1] 1) + climbStairs(n [1] 2);

Drag options to blanks, or click blank then click option'

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using + instead of - causing incorrect recursion.

Using * or / which are not valid here.

We subtract 1 and 2 from n to find ways to climb remaining steps.

4fill in blank

hard

Fill both blanks to create a loop that fills dp array for climbing stairs.

DSA C

for (int i = 3; i <= n; i++) {
    dp[i] = dp[i [1] 1] + dp[i [2] 2];
}

Drag options to blanks, or click blank then click option'

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition or multiplication instead of subtraction in indices.

Mixing operators causing wrong dp indexing.

We subtract 1 and 2 to get previous dp values to sum.

5fill in blank

hard

Fill all three blanks to initialize dp array and return the final result.

DSA C

int dp[n+1];
dp[1] = [1];
dp[2] = [2];
return dp[[3]];

Drag options to blanks, or click blank then click option'

A1

B2

Cn

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Returning dp[0] or dp[1] instead of dp[n].

Incorrect base values for dp[1] or dp[2].

dp[1] = 1 way, dp[2] = 2 ways, return dp[n] for total ways.

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 number of steps you skip

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

C. The maximum number of steps you can take at once

D. The minimum number of steps to reach the top

Solution

1. Step 1: Understand the problem statement

   The problem asks for the count of all possible ways to climb to the top using steps of size 1 or 2.

2. Step 2: Identify what is being counted

   We count distinct 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 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 step size instead of counting ways
• Ignoring the step size options

2. Which of the following is the correct base case initialization for the Climbing Stairs problem in C?

easy

A. int ways[1] = 1; int ways[2] = 2;

B. int ways[0] = 1; int ways[1] = 2;

C. int ways[0] = 0; int ways[1] = 1;

D. int ways[1] = 0; int ways[2] = 1;

Solution

1. Step 1: Identify base cases for climbing stairs

   For 1 step, ways = 1 (only one way). For 2 steps, ways = 2 (1+1 or 2).

2. Step 2: Match base cases with options

   int ways[1] = 1; int ways[2] = 2; correctly sets ways[1] = 1 and ways[2] = 2.

3. Final Answer:

   int ways[1] = 1; int ways[2] = 2; -> Option A

4. Quick Check:

   Base cases: ways[1]=1, ways[2]=2 [OK]

Hint: Remember ways(1)=1 and ways(2)=2 base cases [OK]

Common Mistakes:

• Setting ways[0] incorrectly
• Confusing ways for 1 and 2 steps
• Using zero for ways[1]

3. What is the output of the following C code snippet for n = 4?

int climbStairs(int n) {
    if (n <= 2) return n;
    int a = 1, b = 2, c;
    for (int i = 3; i <= n; i++) {
        c = a + b;
        a = b;
        b = c;
    }
    return b;
}

printf("%d", climbStairs(4));

medium

A. 7

B. 6

C. 5

D. 4

Solution

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

   Initial: a=1, b=2 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 b=5, which is the number of ways to climb 4 steps.

3. Final Answer:

   5 -> Option C

4. Quick Check:

   Ways(4) = 5 [OK]

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

Common Mistakes:

• Returning a instead of b
• Off-by-one errors in loop
• Incorrect initial values

4. Identify the bug in this C function for climbing stairs:

int climbStairs(int n) {
    int ways[n+1];
    ways[1] = 1;
    ways[2] = 2;
    for (int i = 3; i <= n; i++) {
        ways[i] = ways[i-1] + ways[i-2];
    }
    return ways[n];
}

medium

A. Array ways is not initialized for ways[0]

B. No bug, code works correctly

C. Accessing ways[2] before initialization

D. Loop should start from i = 2

Solution

1. Step 1: Check array initialization

   Array ways is declared but ways[0] is not initialized, which may cause undefined behavior if accessed.

2. Step 2: Analyze usage of ways[0]

   Though ways[0] is not used in the loop, some compilers or logic might expect it initialized; best practice is to initialize ways[0] = 0.

3. Final Answer:

   Array ways is not initialized for ways[0] -> Option A

4. Quick Check:

   Uninitialized ways[0] can cause issues [OK]

Hint: Always initialize all array elements before use [OK]

Common Mistakes:

• Ignoring ways[0] initialization
• Assuming no undefined behavior without ways[0]
• Starting loop incorrectly

5. You want to find the number of ways to climb 10 steps but can take 1, 2, or 3 steps at a time. Which modification to the standard climbing stairs approach is correct?

hard

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

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

C. Use ways[i] = ways[i-1] + ways[i-2] with base cases ways[0]=1, ways[1]=1

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

Solution

1. Step 1: Understand step size options

   Since steps can be 1, 2, or 3, ways to reach step i include ways to reach i-1, i-2, and i-3.

2. Step 2: Formulate recurrence and base cases

   ways[i] = ways[i-1] + ways[i-2] + ways[i-3]; base cases: ways[0]=1 (one way to stand still), ways[1]=1, ways[2]=2.

3. Final Answer:

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

4. Quick Check:

   Sum previous 3 ways for steps 1,2,3 [OK]

Hint: Add ways for all allowed step sizes [OK]

Common Mistakes:

• Multiplying ways instead of adding
• Ignoring the 3-step option
• Using wrong base cases