Array Traversal Patterns in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Nested Loop Traversal

What is the output of the following C# code that traverses a 2D array using nested loops?

DSA C#

int[,] matrix = { {1, 2}, {3, 4} };
for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
        Console.Write(matrix[i, j] + " ");
    }
}

A 1 2 3 4

B 1 3 2 4

C 4 3 2 1

D 2 1 4 3

💡 Hint

Remember that the outer loop controls rows and the inner loop controls columns.

The outer loop iterates over rows (i), and the inner loop iterates over columns (j). So the elements are printed row-wise: 1 2 3 4.

❓ Predict Output

intermediate

Output of Reverse Array Traversal

What is the output of this C# code that traverses an array in reverse order?

DSA C#

int[] arr = {5, 10, 15, 20};
for (int i = arr.Length - 1; i >= 0; i--) {
    Console.Write(arr[i] + " ");
}

A 15 20 10 5

B 5 10 15 20

C 20 10 15 5

D 20 15 10 5

💡 Hint

The loop starts from the last index and moves backward.

The loop prints elements starting from the last index to the first, so output is 20 15 10 5.

🧠 Conceptual

advanced

Understanding Zigzag Traversal Pattern

In a zigzag traversal of a 2D array, rows are traversed alternately from left to right and right to left. Given the 2D array below, what is the output of zigzag traversal?

DSA C#

int[,] matrix = { {1, 2, 3}, {4, 5, 6}, {7, 8, 9} };
// Zigzag traversal code not shown

A 1 2 3 4 5 6 9 8 7

B 1 2 3 6 5 4 9 8 7

C 1 2 3 6 5 4 7 8 9

D 1 2 3 4 5 6 7 8 9

💡 Hint

Odd rows are traversed right to left, even rows left to right.

Row 0 left to right: 1 2 3; Row 1 right to left: 6 5 4; Row 2 left to right: 7 8 9. So output is 1 2 3 6 5 4 7 8 9 (option C).

🔧 Debug

advanced

Identify the Error in Diagonal Traversal Code

What error will this C# code produce when trying to print the main diagonal of a 2D square array?

DSA C#

int[,] matrix = { {1, 2}, {3, 4} };
for (int i = 0; i < matrix.GetLength(0); i++) {
    Console.Write(matrix[i, i] + " ");
}

A Prints: 1 4

B IndexOutOfRangeException

C No output, infinite loop

D SyntaxError

💡 Hint

Check the loop boundary condition carefully.

matrix.GetLength(0) returns 2, valid indices are 0 and 1. Loop runs i=0 to 1 inclusive, so it prints 1 and 4 without error.

🚀 Application

expert

Count Unique Elements Using Array Traversal

Given an integer array, which code snippet correctly counts the number of unique elements using array traversal?

A

int count = 0;
for (int i = 0; i < arr.Length; i++) {
    if (arr[i] != arr[i + 1]) count++;
}
Console.WriteLine(count);

B

int count = 0;
for (int i = 0; i < arr.Length; i++) {
    bool isUnique = true;
    for (int j = 0; j < i; j++) {
        if (arr[i] == arr[j]) {
            isUnique = false;
            break;
        }
    }
    if (isUnique) count++;
}
Console.WriteLine(count);

C

int count = 0;
for (int i = 0; i < arr.Length; i++) {
    for (int j = i + 1; j < arr.Length; j++) {
        if (arr[i] == arr[j]) count++;
    }
}
Console.WriteLine(count);

D

int count = 0;
HashSet<int> set = new HashSet<int>();
for (int i = 0; i < arr.Length; i++) {
    set.Add(arr[i]);
}
count = set.Count;
Console.WriteLine(count);

💡 Hint

Check which code counts unique elements correctly without errors.

Option B correctly checks each element against previous elements to count unique ones. Option B counts duplicates, option B causes IndexOutOfRangeException, option B uses HashSet which is not allowed as per traversal-only constraint.

Practice

1. What does the following C# code do?
int[] arr = {1, 2, 3, 4};
for (int i = 0; i < arr.Length; i++) {
Console.Write(arr[i] + " ");
}

easy

A. Prints all elements of the array from start to end

B. Prints all elements of the array in reverse order

C. Prints only the first element of the array

D. Causes a runtime error due to wrong loop condition

Solution

1. Step 1: Understand the loop range

   The loop starts at 0 and runs while i is less than arr.Length (4), so i goes 0,1,2,3.

2. Step 2: Print elements in order

   Each iteration prints arr[i], so it prints 1 2 3 4 in order.

3. Final Answer:

   Prints all elements of the array from start to end -> Option A

4. Quick Check:

   Loop from 0 to length-1 prints all elements [OK]

Hint: Loop from 0 to length-1 prints all elements in order [OK]

Common Mistakes:

• Thinking loop prints in reverse
• Confusing loop condition with <= instead of <
• Assuming only first element prints

2. Which of the following is the correct syntax to traverse an array arr in C# using a foreach loop?

easy

A. foreach (int i in arr) { Console.Write(i); }

B. foreach (arr in int i) { Console.Write(i); }

C. foreach (int arr in i) { Console.Write(i); }

D. foreach (int i; arr) { Console.Write(i); }

Solution

1. Step 1: Recall foreach syntax

   The correct syntax is: foreach (type variable in collection) { ... }

2. Step 2: Match syntax to options

   foreach (int i in arr) { Console.Write(i); } matches correct syntax: foreach (int i in arr) { Console.Write(i); }

3. Final Answer:

   foreach (int i in arr) { Console.Write(i); } -> Option A

4. Quick Check:

   foreach loops use 'variable in collection' syntax [OK]

Hint: Remember foreach uses 'variable in collection' syntax [OK]

Common Mistakes:

• Swapping variable and collection order
• Using semicolon instead of 'in'
• Incorrect variable declaration inside foreach

3. What is the output of the following code?

int[] arr = {10, 20, 30, 40};
for (int i = arr.Length - 1; i >= 0; i--) {
  Console.Write(arr[i] + " ");
}

medium

A. 40 30 20

B. 10 20 30 40

C. 40 30 20 10

D. Runtime error due to loop condition

Solution

1. Step 1: Analyze loop start and end

   Loop starts at arr.Length - 1 = 3 and runs while i >= 0, decrementing i each time.

2. Step 2: Trace printed elements

   Prints arr[3]=40, arr[2]=30, arr[1]=20, arr[0]=10 in that order.

3. Final Answer:

   40 30 20 10 -> Option C

4. Quick Check:

   Loop from end to start prints reversed array [OK]

Hint: Loop from length-1 down to 0 prints array in reverse [OK]

Common Mistakes:

• Confusing loop direction and printing forward
• Off-by-one errors in loop condition
• Assuming runtime error due to negative index

4. Find the error in this array traversal code:

int[] arr = {5, 10, 15};
for (int i = 0; i <= arr.Length; i++) {
  Console.Write(arr[i] + " ");
}

medium

A. Array declaration is incorrect

B. Variable i should start from 1, not 0

C. Console.Write should be Console.WriteLine

D. Loop condition should be i < arr.Length, not <= arr.Length

Solution

1. Step 1: Check loop condition

   Loop runs while i <= arr.Length, but arr indices go from 0 to arr.Length - 1.

2. Step 2: Identify out-of-range access

   When i == arr.Length (3), arr[3] is accessed causing runtime error (IndexOutOfRangeException).

3. Final Answer:

   Loop condition should be i < arr.Length, not <= arr.Length -> Option D

4. Quick Check:

   Array indices go 0 to length-1, use < not <= [OK]

Hint: Use i < arr.Length to avoid out-of-range errors [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Starting loop at 1 instead of 0
• Confusing Write and WriteLine methods

5. Given an integer array arr, which code snippet correctly prints only the even-indexed elements (0, 2, 4, ...) of the array?

hard

A. foreach (int i in arr) { if (i % 2 == 0) Console.Write(i + " "); }

B. for (int i = 0; i < arr.Length; i += 2) { Console.Write(arr[i] + " "); }

C. for (int i = 1; i < arr.Length; i += 2) { Console.Write(arr[i] + " "); }

D. for (int i = 0; i < arr.Length; i++) { if (arr[i] % 2 == 0) Console.Write(arr[i] + " "); }

Solution

1. Step 1: Understand requirement

   We want to print elements at even indexes: 0, 2, 4, ...

2. Step 2: Analyze options

   for (int i = 0; i < arr.Length; i += 2) { Console.Write(arr[i] + " "); } increments i by 2 starting at 0, so it accesses even indexes correctly.
   for (int i = 1; i < arr.Length; i += 2) { Console.Write(arr[i] + " "); } accesses odd indexes.
   foreach (int i in arr) { if (i % 2 == 0) Console.Write(i + " "); } checks element values for evenness, not indexes.
   for (int i = 0; i < arr.Length; i++) { if (arr[i] % 2 == 0) Console.Write(arr[i] + " "); } prints elements with even values, not indexes.

3. Final Answer:

   for (int i = 0; i < arr.Length; i += 2) { Console.Write(arr[i] + " "); } -> Option B

4. Quick Check:

   Increment index by 2 from 0 to get even indexes [OK]

Hint: Increment loop index by 2 starting at 0 for even indexes [OK]

Common Mistakes:

• Confusing even index with even element value
• Starting loop at 1 for even indexes
• Using foreach without index tracking