Left Rotate Array by One Algorithm

#include <iostream>

void leftRotateByOne(int arr[], int n) {
    int temp = arr[0];
    for (int i = 0; i < n - 1; i++) {
        arr[i] = arr[i + 1];
    }
    arr[n - 1] = temp;
}

int main() {
    const int size = 5;  // Change the array size as needed
    int arr[size] = {1, 2, 3, 4, 5};  // Initialize the array

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

    leftRotateByOne(arr, size);

    std::cout << "\nArray after Left Rotation by One: ";
    for (int i = 0; i < size; i++) {
        std::cout << arr[i] << " ";
    }

    return 0;
}

Algorithm Logic:

1. Store the first element of the array in a temporary variable (temp).

2. Shift all elements of the array to the left by one position.

3. Assign the value of temp to the last element of the array.

Dry Run:

Suppose the initial array is [1, 2, 3, 4, 5].

1. temp = 1

2. Shift elements: [2, 3, 4, 5, 5]

3. Assign temp to the last element: [2, 3, 4, 5, 1]

# Left Rotate Array by One

This C++ program demonstrates left rotation of an array by one place.

## How it Works

1. The program initializes an array.
2. It rotates the array to the left by one position.
3. The rotated array is then printed.

## Example

Original Array: 1 2 3 4 5 Array after Left Rotation by One: 2 3 4 5 1