Array Traversal Patterns in DSA C++ - Build from Scratch

Array Traversal Patterns

📖 Scenario: You are helping a small grocery store manage their daily sales data. The store records the number of items sold each day in an array. You want to analyze this data by visiting each day's sales count one by one.

🎯 Goal: Build a simple C++ program that creates an array of daily sales, sets up a counter for total sales, traverses the array to sum all sales, and finally prints the total sales.

📋 What You'll Learn

Create an integer array named dailySales with exactly these values: 5, 8, 12, 3, 7

Create an integer variable named totalSales and initialize it to 0

Use a for loop with an integer variable i to traverse the dailySales array from start to end

Inside the loop, add the current day's sales to totalSales

Print the value of totalSales after the loop finishes

💡 Why This Matters

🌍 Real World

Stores and businesses often keep daily sales data in arrays or lists. Traversing these arrays helps calculate totals, averages, or other statistics to understand performance.

💼 Career

Understanding array traversal is fundamental for software developers working with data processing, reporting, and analytics in many industries.

1

Create the daily sales array

Create an integer array called dailySales with these exact values: 5, 8, 12, 3, 7

DSA C++

# Your code here

Hint

Use C++ array syntax with size 5 and initialize it with the given numbers inside curly braces.

2

Initialize total sales counter

Create an integer variable called totalSales and set it to 0

DSA C++

int dailySales[5] = {5, 8, 12, 3, 7};
# Your code here

Hint

Declare an integer variable and assign it zero to start counting total sales.

3

Traverse the array and sum sales

Use a for loop with an integer variable i to traverse the dailySales array from index 0 to 4. Inside the loop, add dailySales[i] to totalSales

DSA C++

int dailySales[5] = {5, 8, 12, 3, 7};
int totalSales = 0;
# Your code here

Hint

Use a for loop starting at 0, ending before 5, and add each element to totalSales inside the loop.

4

Print the total sales

Write a std::cout statement to print the value of totalSales

DSA C++

int dailySales[5] = {5, 8, 12, 3, 7};
int totalSales = 0;
for (int i = 0; i < 5; i++) {
    totalSales += dailySales[i];
}
# Your code here

Hint

Use std::cout << totalSales << std::endl; to print the total sales.

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