Array Traversal Patterns in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print all elements of the array using a for loop.

DSA C++

int arr[] = {1, 2, 3, 4, 5};
int n = 5;
for (int i = 0; i [1] n; i++) {
    std::cout << arr[i] << " ";
}

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

A<

B>

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using i <= n causes out-of-bounds access.

Using i > n or i >= n causes the loop to never run.

The loop should run from 0 up to but not including n, so i < n is correct.

2fill in blank

medium

Complete the code to traverse the array backwards and print elements.

DSA C++

int arr[] = {10, 20, 30, 40, 50};
int n = 5;
for (int i = n - 1; i [1] 0; i--) {
    std::cout << arr[i] << " ";
}

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

A>

B>=

C<

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using i < 0 causes the loop to never run.

Using i <= 0 stops the loop too early.

The loop starts at n-1 and decrements until i >= 0 to include index 0.

3fill in blank

hard

Fix the error in the code to correctly print every second element of the array.

DSA C++

int arr[] = {5, 10, 15, 20, 25, 30};
int n = 6;
for (int i = 0; i [1] n; i += 2) {
    std::cout << arr[i] << " ";
}

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

A<=

B>

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using i <= n causes out-of-bounds access.

Using i > n or i >= n causes the loop to never run.

To avoid out-of-bounds access, the loop should run while i is less than n, so i < n is correct.

4fill in blank

hard

Fill both blanks to create a loop that prints elements at odd indices.

DSA C++

int arr[] = {2, 4, 6, 8, 10, 12};
int n = 6;
for (int i = [1]; i [2] n; i += 2) {
    std::cout << arr[i] << " ";
}

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

A1

B0

C<=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Starting at 0 prints even indices.

Using i <= n causes out-of-bounds access.

Start at index 1 to get odd indices, and loop while i < n to avoid out-of-bounds.

5fill in blank

hard

Fill all three blanks to create a loop that prints elements divisible by 3.

DSA C++

int arr[] = {3, 5, 6, 7, 9, 12};
int n = 6;
for (int i = 0; i [1] n; i++) {
    if (arr[i] [2] 3 [3] 0) {
        std::cout << arr[i] << " ";
    }
}

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

A<=

B<

C%

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using i <= n causes out-of-bounds access.

Using == instead of % causes syntax errors.

Checking for non-zero remainder instead of zero.

Loop while i < n to stay in bounds, use % to check remainder, and == 0 to check divisibility.

Practice

1. What does the following loop do when traversing an array arr of size n?

for (int i = 0; i < n; i++) { /* code */ }

easy

A. It visits each element from the first to the last in order.

B. It visits elements in reverse order from last to first.

C. It visits only even-indexed elements.

D. It skips the first element and visits the rest.

Solution

1. Step 1: Understand the loop variable and range

   The loop variable i starts at 0 and increases by 1 until it reaches n-1.

2. Step 2: Determine the traversal order

   Since i increments from 0 to n-1, the loop visits elements from the first to the last in order.

3. Final Answer:

   It visits each element from the first to the last in order. -> Option A

4. Quick Check:

   Forward traversal = It visits each element from the first to the last in order. [OK]

Hint: Loop from 0 to n-1 means forward traversal [OK]

Common Mistakes:

• Thinking it visits elements in reverse order
• Assuming it skips elements
• Confusing index start with 1 instead of 0

2. Which of the following is the correct syntax to traverse an array arr of size n backwards in C++?

easy

A. for (int i = n-1; i >= 0; i--) { /* code */ }

B. for (int i = 0; i < n; i++) { /* code */ }

C. for (int i = n; i > 0; i--) { /* code */ }

D. for (int i = n-1; i <= 0; i--) { /* code */ }

Solution

1. Step 1: Identify correct start and end conditions

   To traverse backwards, start at n-1 (last index) and go down to 0.

2. Step 2: Check loop condition and decrement

   The loop must continue while i >= 0 and decrement i-- each time.

3. Final Answer:

   for (int i = n-1; i >= 0; i--) { /* code */ } -> Option A

4. Quick Check:

   Backward loop: start=n-1, end=0, decrement [OK]

Hint: Backward loop starts at n-1 and decrements to 0 [OK]

Common Mistakes:

• Using i <= 0 instead of i >= 0
• Starting at n instead of n-1
• Incrementing instead of decrementing

3. What is the output of the following C++ code?

int arr[] = {10, 20, 30, 40, 50};
int n = 5;
for (int i = 1; i < n; i += 2) {
    std::cout << arr[i] << " ";
}

medium

A. 10 30 50

B. 20 40 50

C. 10 20 30 40 50

D. 20 40

Solution

1. Step 1: Identify loop indices visited

   The loop starts at i=1 and increments by 2: i=1, 3, then stops before 5.

2. Step 2: Find elements at these indices

   arr[1] = 20, arr[3] = 40. These are printed separated by space.

3. Final Answer:

   20 40 -> Option D

4. Quick Check:

   Indices 1 and 3 values = 20 40 [OK]

Hint: Step by 2 from 1 visits odd indices [OK]

Common Mistakes:

• Including index 5 which is out of bounds
• Starting from 0 instead of 1
• Printing all elements instead of skipping

4. Find the error in the following code snippet that tries to print all elements of an array arr of size n:

for (int i = 0; i <= n; i++) {
    std::cout << arr[i] << " ";
}

medium

A. The loop should increment i-- instead of i++.

B. The loop condition should be i < n instead of i <= n.

C. The array index should be i-1 inside the loop.

D. The loop variable i should start from 1.

Solution

1. Step 1: Analyze loop boundary condition

   The loop runs while i <= n, so it accesses arr[n] which is out of bounds.

2. Step 2: Correct the loop condition

   Change condition to i < n to access valid indices 0 to n-1.

3. Final Answer:

   The loop condition should be i < n instead of i <= n. -> Option B

4. Quick Check:

   Array indices 0 to n-1 valid, not n [OK]

Hint: Array indices go from 0 to n-1, not n [OK]

Common Mistakes:

• Using <= n causes out-of-bounds access
• Starting loop at 1 misses first element
• Decrementing i in a forward loop

5. Given an array arr of size n, which code snippet correctly prints elements at even indices in reverse order?

hard

A. for (int i = n-1; i >= 0; i -= 2) { std::cout << arr[i] << " "; }

B. for (int i = 0; i < n; i += 2) { std::cout << arr[i] << " "; }

C. for (int i = (n % 2 == 0 ? n-2 : n-1); i >= 0; i -= 2) { std::cout << arr[i] << " "; }

D. for (int i = n-2; i <= 0; i -= 2) { std::cout << arr[i] << " "; }

Solution

1. Step 1: Determine the highest even index to start

   If n is even, highest even index is n-2; if odd, it's n-1.

2. Step 2: Loop backwards by 2 from that index down to 0

   The loop decrements i by 2 each time, printing even indices in reverse order.

3. Final Answer:

   for (int i = (n % 2 == 0 ? n-2 : n-1); i >= 0; i -= 2) { std::cout << arr[i] << " "; } -> Option C

4. Quick Check:

   Start at highest even index, decrement by 2 [OK]

Hint: Start at highest even index, decrement by 2 [OK]

Common Mistakes:

• Starting at n-1 always (may be odd index)
• Loop condition incorrect (i <= 0 instead of i >= 0)
• Incrementing instead of decrementing in reverse