Left Rotation - HackerRank | C++ Implementation
A left rotation operation on an array of size n shifts each of the array’s elements 1 unit to the left. For example, if 2 left rotations are performed on array [1, 2, 3, 4, 5], then the array would become [3, 4, 5, 1, 2].
Given an array of n integers and a number, d, perform d left rotations on the array. Then print the updated array as a single line of space-separated integers.
Read the full problem here : Left Rotation
This is a very easy problem. In the above figure we see that after two rotations, the element at index 2 is the first element in the new array. Thus after d rotations, the element at index d is the first element. And all elements after dth elements with follow.
for (std::size_t i = rotations; i < arr.size(); ++i)
{
rotated_array.push_back(arr[i]);
}
Now, elements from 0 to d-1 indices are added to the new rotated array.
for (std::size_t i = 0; i < rotations; ++i)
{
rotated_array.push_back(arr[i]);
}
#include <iostream>
#include <vector>
std::vector<int> left_rotation(std::vector<int>& arr, int rotations)
{
std::vector<int> rotated_array;
for (std::size_t i = rotations; i < arr.size(); ++i)
{
rotated_array.push_back(arr[i]);
}
for (std::size_t i = 0; i < rotations; ++i)
{
rotated_array.push_back(arr[i]);
}
return rotated_array;
}
int main()
{
int num_elements, num_left_rotations;
std::cin >> num_elements;
std::cin >> num_left_rotations;
std::vector<int> input_array(num_elements);
for (int i = 0; i < num_elements; ++i)
{
std::cin >> input_array[i];
}
std::vector<int> result_array = left_rotation(input_array, num_left_rotations);
for (int i = 0; i < num_elements; ++i)
{
std::cout << result_array[i] <<" ";
}
std::cout << "\n";
}
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