public static List<Integer> rotateLeft(int d, List<Integer> arr) { int n = arr.size(); List<Integer> result = new ArrayList<>(n); for (int i = 0; i < n; i++) { int newIndex = (i + n - d) % n; result.add(arr.get(newIndex)); } return result; }}

public ListNode rotateRight(ListNode head, int k) { ListNode slow = head; ListNode fast = head; for(int i=0; i<k;i++){ fast=fast.next; }

A left rotation operation on an array of size shifts each of the array's elements unit to the left. Given an integer, , rotate the array that many steps left and return the result.ExampleAfter rotations, .Function DescriptionComplete the rotateLeft function in the editor below.rotateLeft has the following parameters:int d: the amount to rotate byint arr[n]: the array to rotateReturnsint[n]: the rotated arrayInput FormatThe first line contains two space-separated integers that denote , the number of integers, and , the number of left rotations to perform.The second line contains space-separated integers that describe .ConstraintsSample Input5 41 2 3 4 5Sample Output5 1 2 3 4ExplanationTo perform left rotations, the array undergoes the following sequence of changes:

#include <iostream>#include <vector>void rotateMatrix(int** matrix, int n) { // Transpose the matrix for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { std::swap(matrix[i][j], matrix[j][i]); } } // Reverse each row for (int i = 0; i < n; i++) { int left = 0; int right = n - 1; while (left < right) { std::swap(matrix[i][left], matrix[i][right]); left++; right--; } }}void printMatrix(int** matrix, int n) { for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { std::cout << matrix[i][j] << ' '; } std::cout << std::endl; }}int main() { int n; std::cin >> n; if (n <= 0) { std::cout << "Size of the matrix should be positive!" << std::endl; return 0; } // Allocate dynamic memory for the matrix int** matrix = new int*[n]; for (int i = 0; i < n; i++) { matrix[i] = new int[n]; } // Read the matrix elements for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { std::cin >> matrix[i][j]; } } std::cout << "Original Matrix:" << std::endl; printMatrix(matrix, n); rotateMatrix(matrix, n); std::cout << "Matrix after 90-degree clockwise rotation:" << std::endl; printMatrix(matrix, n); // Deallocate the dynamic memory for (int i = 0; i < n; i++) { delete[] matrix[i]; } delete[] matrix; return 0;}

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:[4,5,6,7,0,1,2] if it was rotated 4 times.[0,1,2,4,5,6,7] if it was rotated 7 times.Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].Given the sorted rotated array nums of unique elements, return the minimum element of this array.You must write an algorithm that runs in O(log n) time.

Phương thức nào sau đây cho phép xoay vòng các phần tử trong ArrayList? void rotate (List list, int distance) void reverse (List list) void swap(List list, int i, int j) void sort (List list)