Algorithm

First, we will perform (d%n) to get the effective number of rotations.
We will store the first d elements in a temporary array.
We will shift the last (n-d) elements by d places to the left using a loop(say i) that will start from index d and run up to index n-1 (in case of 0-based indexing). Inside the loop, the shifting will be like: arr[i-d] = arr[i]..
After that, we will place the elements of the temporary array in the last d places of the given array using another loop that starts from (n-d) and runs up to n-1 (0-based index).
While selecting elements from the temporary array, one of the ways is to use a variable(say j and starts from 0) and increment it inside the loop.
The efficient way will be the following. The element at index i in the temporary array will be placed at index (n-d+i) in the given array. So, to place the elements in the correct position, inside the loop, we will do this: arr[i] = temp[i-(n-d)].

Assume the given array is {1,2,3,4,5,6,7} and d = 3.

Code

#include <iostream>
using namespace std;

class LeftRotation {
public:
    // Function to rotate the array to the left by d positions
    void leftRotate(int arr[], int n, int d) {
        i