Array Traversal Patterns in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Nested Loop Array Traversal

What is the printed output of the following C code snippet that traverses a 2D array?

DSA C

int arr[2][3] = {{1, 2, 3}, {4, 5, 6}};
for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 3; j++) {
        printf("%d ", arr[i][j]);
    }
    printf("\n");
}

A 1 2 3 4 5 6 \n

B 4 5 6 \n1 2 3 \n

C 1 4 2 \n3 5 6 \n

D 1 2 3 \n4 5 6 \n

💡 Hint

Think about how the outer loop controls rows and the inner loop controls columns.

The outer loop runs twice for each row, and the inner loop prints all columns in that row before moving to the next line.

❓ Predict Output

intermediate

Output of Single Loop Traversal with Step

What is the output of this C code that traverses an array with a step of 2?

DSA C

int arr[] = {10, 20, 30, 40, 50, 60};
int n = 6;
for (int i = 0; i < n; i += 2) {
    printf("%d ", arr[i]);
}

A 10 30 50

B 20 40 60

C 10 20 30 40 50 60

D 60 50 40 30 20 10

💡 Hint

The loop increments by 2, so it skips every other element.

Starting at index 0, the loop prints elements at indices 0, 2, and 4.

🔧 Debug

advanced

Identify the Error in Reverse Array Traversal

What error does this C code produce when trying to print an array in reverse?

DSA C

int arr[] = {1, 2, 3, 4, 5};
int n = 5;
for (int i = n - 1; i >= 0; i--) {
    printf("%d ", arr[i]);
}

A Prints: 5 4 3 2 1

B Compilation error: invalid loop condition

C Prints: 5 4 3 2 1 0 (out of bounds access)

D Runtime error: Array index out of bounds

💡 Hint

Check the starting index and valid array indices.

Array indices go from 0 to n-1. Starting at n accesses arr[5], which is out of bounds causing runtime error.

🧠 Conceptual

advanced

Understanding Traversal Order in Zigzag Pattern

Given a 2D array, which traversal pattern prints elements in a zigzag order (left to right on even rows, right to left on odd rows)?

DSA C

int arr[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
for (int i = 0; i < 3; i++) {
    if (i % 2 == 0) {
        for (int j = 0; j < 3; j++) {
            printf("%d ", arr[i][j]);
        }
    } else {
        for (int j = 2; j >= 0; j--) {
            printf("%d ", arr[i][j]);
        }
    }
    printf("\n");
}

A 3 2 1 \n4 5 6 \n9 8 7 \n

B 1 2 3 \n4 5 6 \n7 8 9 \n

C 1 2 3 \n6 5 4 \n7 8 9 \n

D 1 3 2 \n4 6 5 \n7 9 8 \n

💡 Hint

Even rows go left to right, odd rows go right to left.

The code prints row 0 left to right, row 1 right to left, row 2 left to right, creating a zigzag pattern.

🚀 Application

expert

Count Unique Elements Using Array Traversal

Given an unsorted integer array, which code snippet correctly counts the number of unique elements using only array traversal and no extra data structures?

DSA C

int arr[] = {2, 3, 2, 4, 3, 5};
int n = 6;
int count = 0;
for (int i = 0; i < n; i++) {
    int j;
    for (j = 0; j < i; j++) {
        if (arr[i] == arr[j]) {
            break;
        }
    }
    if (j == i) {
        count++;
    }
}
printf("%d", count);

A 5

B 4

C 3

D 6

💡 Hint

Count increments only when current element is not found before in the array.

The code checks if the current element appeared before; if not, it counts it as unique. The unique elements are 2, 3, 4, 5 totaling 4.