Array Rotation Techniques in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Left Rotation by 2

What is the output of the array after performing a left rotation by 2 positions?

DSA C

int arr[] = {10, 20, 30, 40, 50};
int n = 5, d = 2;
// Left rotate by d
int rotated[5];
for(int i = 0; i < n; i++) {
    rotated[i] = arr[(i + d) % n];
}
for(int i = 0; i < n; i++) {
    printf("%d ", rotated[i]);
}

A 20 30 40 50 10

B 40 50 10 20 30

C 30 40 50 10 20

D 50 10 20 30 40

💡 Hint

Think about moving the first two elements to the end.

Left rotation by 2 moves elements at index 0 and 1 to the end, so the array starts from index 2.

❓ Predict Output

intermediate

Output of Right Rotation by 3

What is the output of the array after performing a right rotation by 3 positions?

DSA C

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

A 6 1 2 3 4 5

B 4 5 6 1 2 3

C 3 4 5 6 1 2

D 5 6 1 2 3 4

💡 Hint

Right rotation by 3 moves last 3 elements to the front.

Right rotation by 3 shifts elements so that the last 3 elements come to the front.

🔧 Debug

advanced

Identify the Error in Rotation Code

What error will this code produce when trying to left rotate an array by d positions?

DSA C

int arr[] = {7, 8, 9, 10};
int n = 4, d = 5;
int rotated[4];
for(int i = 0; i < n; i++) {
    rotated[i] = arr[(i + d) % n];
}
for(int i = 0; i < n; i++) {
    printf("%d ", rotated[i]);
}

A No error, output: 8 9 10 7

B Array index out of bounds error

C Output: 7 8 9 10

D Compilation error due to invalid syntax

💡 Hint

Consider how modulo operator handles d greater than n.

Modulo operator ensures index wraps correctly even if d > n, so no error occurs.

❓ Predict Output

advanced

Output of Reversal Algorithm for Left Rotation

What is the output of the array after left rotating by 3 using the reversal algorithm?

DSA C

void reverse(int arr[], int start, int end) {
    while(start < end) {
        int temp = arr[start];
        arr[start] = arr[end];
        arr[end] = temp;
        start++;
        end--;
    }
}

int arr[] = {1, 2, 3, 4, 5, 6, 7};
int n = 7, d = 3;
reverse(arr, 0, d - 1);
reverse(arr, d, n - 1);
reverse(arr, 0, n - 1);
for(int i = 0; i < n; i++) {
    printf("%d ", arr[i]);
}

A 4 5 6 7 1 2 3

B 7 6 5 4 3 2 1

C 5 6 7 1 2 3 4

D 3 2 1 7 6 5 4

💡 Hint

Reversal algorithm reverses parts then whole array.

First reverse first 3 elements, then last 4, then whole array to get left rotation by 3.

🧠 Conceptual

expert

Minimum Number of Rotations to Restore Original Array

Given an array of length n, what is the minimum number of left rotations needed to restore the array to its original order?

A 0

B 1

C n - 1

D n

💡 Hint

Think about how many times you must rotate to get back to start.

Rotating n times brings every element back to its original position.