Juggling Algorithm for Array Rotation
Last Updated :
09 Nov, 2024
Given an array arr[], the task is to rotate the array by d positions to the left using juggling algorithm.
Examples:
Input: arr[] = {1, 2, 3, 4, 5, 6}, d = 2
Output: {3, 4, 5, 6, 1, 2}
Explanation: After first left rotation, arr[] becomes {2, 3, 4, 5, 6, 1} and after the second rotation, arr[] becomes {3, 4, 5, 6, 1, 2}
Input: arr[] = {1, 2, 3}, d = 4
Output: {2, 3, 1}
Explanation: The array is rotated as follows:
- After first left rotation, arr[] = {2, 3, 1}
- After second left rotation, arr[] = {3, 1, 2}
- After third left rotation, arr[] = {1, 2, 3}
- After fourth left rotation, arr[] = {2, 3, 1}
Approach:
The idea behind Juggling Algorithm is that we can rotate all elements of array using cycles. Each cycle is independent and represents a group of elements that will shift among themselves during the rotation. If the starting index of a cycle is i, then next elements of the cycle will be present at indices (i + d) % n, (i + 2d) % n, (i + 3d) % n ... and so on till we reach back to index i.
So for any index i, we know that after rotation we will have arr[(i + d) % n] at index i. Now, for every index in the cycle, we will place the element which should be present at that index after the array is rotated.
How many independent cycles will be there if an array of size n is rotated by d positions?
If an array of size n is rotated by d places, then the number of independent cycles will be gcd(n, d). When we rotate the array, the cycle increments in steps of d and terminates when we reach the starting point. This means that the distance travelled in each cycle will be a multiple of n, which is defined as lcm(n, d). So, number of elements in each cycle = lcm(n, d)/d. Therefore, to cover all n elements, the total number of cycles will be n/lcm(n, d) = gcd(n, d).
Below is the implementation of the algorithm:
C++
// C++ Program to left rotate the array by d positions
// using Juggling Algorithm
#include <bits/stdc++.h>
using namespace std;
// Function to rotate vector
void rotateArr(vector<int> &arr, int d) {
int n = arr.size();
// Handle the case where d > size of array
d %= n;
// Calculate the number of cycles in the rotation
int cycles = __gcd(n, d);
// Process each cycle
for (int i = 0; i < cycles; i++) {
// Start element of current cycle
int startEle = arr[i];
// Start index of current cycle
int currIdx = i, nextIdx;
// Rotate elements till we reach the start of cycle
while(true) {
nextIdx = (currIdx + d) % n;
if(nextIdx == i)
break;
// Update the next index with the current element
arr[currIdx] = arr[nextIdx];
currIdx = nextIdx;
}
// Copy the start element of current cycle at the last
// index of the cycle
arr[currIdx] = startEle;
}
}
int main()
{
vector<int> arr = {1, 2, 3, 4, 5, 6};
int d = 2;
rotateArr(arr, d);
// Print the rotated array
for (int i = 0; i < arr.size(); i++)
cout << arr[i] << " ";
return 0;
}
// C Program to left rotate the array by d positions
// using Juggling Algorithm
#include <stdio.h>
#include <stdlib.h>
// Function to compute GCD
int gcd(int a, int b) {
while (b != 0) {
int temp = b;
b = a % b;
a = temp;
}
return a;
}
// Function to rotate array
void rotateArr(int arr[], int n, int d) {
// Handle the case where d > size of array
d %= n;
// Calculate the number of cycles in the rotation
int cycles = gcd(n, d);
// Process each cycle
for (int i = 0; i < cycles; i++) {
// Start element of current cycle
int startEle = arr[i];
// Start index of current cycle
int currIdx = i, nextIdx;
// Rotate elements till we reach the start of cycle
while (1) {
nextIdx = (currIdx + d) % n;
if (nextIdx == i)
break;
// Update the next index with the current element
arr[currIdx] = arr[nextIdx];
currIdx = nextIdx;
}
// Copy the start element of current cycle at the last index of the cycle
arr[currIdx] = startEle;
}
}
int main() {
int arr[] = {1, 2, 3, 4, 5, 6};
int n = sizeof(arr) / sizeof(arr[0]);
int d = 2;
rotateArr(arr, n, d);
// Print the rotated array
for (int i = 0; i < n; i++)
printf("%d ", arr[i]);
return 0;
}
// Java Program to left rotate the array by d positions
// using Juggling Algorithm
import java.util.Arrays;
class GfG {
// Function to rotate array
static void rotateArr(int[] arr, int d) {
int n = arr.length;
// Handle the case where d > size of array
d %= n;
// Calculate the number of cycles in the rotation
int cycles = gcd(n, d);
// Process each cycle
for (int i = 0; i < cycles; i++) {
// Start element of current cycle
int startEle = arr[i];
// Start index of current cycle
int currIdx = i, nextIdx;
// Rotate elements till we reach the start of cycle
while (true) {
nextIdx = (currIdx + d) % n;
if (nextIdx == i)
break;
// Update the next index with the current element
arr[currIdx] = arr[nextIdx];
currIdx = nextIdx;
}
// Copy the start element of current cycle at the last
// index of the cycle
arr[currIdx] = startEle;
}
}
// function to compute GCD
static int gcd(int a, int b) {
while (b != 0) {
int temp = b;
b = a % b;
a = temp;
}
return a;
}
public static void main(String[] args) {
int[] arr = {1, 2, 3, 4, 5, 6};
int d = 2;
rotateArr(arr, d);
// Print the rotated array
for (int i = 0; i < arr.length; i++)
System.out.print(arr[i] + " ");
}
}
# Python Program to left rotate the array by d positions
# using Juggling Algorithm
import math
# Function to rotate array
def rotateArr(arr, d):
n = len(arr)
# Handle the case where d > size of array
d %= n
# Calculate the number of cycles in the rotation
cycles = math.gcd(n, d)
# Process each cycle
for i in range(cycles):
# Start element of current cycle
startEle = arr[i]
# Start index of current cycle
currIdx = i
# Rotate elements till we reach the start of cycle
while True:
nextIdx = (currIdx + d) % n
if nextIdx == i:
break
# Update the next index with the current element
arr[currIdx] = arr[nextIdx]
currIdx = nextIdx
# Copy the start element of current cycle at the last
# index of the cycle
arr[currIdx] = startEle
if __name__ == "__main__":
arr = [1, 2, 3, 4, 5, 6]
d = 2
rotateArr(arr, d)
# Print the rotated array
for i in range(len(arr)):
print(arr[i], end=" ")
// C# Program to left rotate the array by d positions
// using Juggling Algorithm
using System;
class GfG {
// Function to rotate array
static void rotateArr(int[] arr, int d) {
int n = arr.Length;
// Handle the case where d > size of array
d %= n;
// Calculate the number of cycles in the rotation
int cycles = gcd(n, d);
// Process each cycle
for (int i = 0; i < cycles; i++) {
// Start element of current cycle
int startEle = arr[i];
// Start index of current cycle
int currIdx = i, nextIdx;
// Rotate elements till we reach the start of cycle
while (true) {
nextIdx = (currIdx + d) % n;
if (nextIdx == i)
break;
// Update the next index with the current element
arr[currIdx] = arr[nextIdx];
currIdx = nextIdx;
}
// Copy the start element of current cycle at the last
// index of the cycle
arr[currIdx] = startEle;
}
}
// function to compute GCD
static int gcd(int a, int b) {
while (b != 0) {
int temp = b;
b = a % b;
a = temp;
}
return a;
}
static void Main(string[] args) {
int[] arr = { 1, 2, 3, 4, 5, 6 };
int d = 2;
rotateArr(arr, d);
// Print the rotated array
for (int i = 0; i < arr.Length; i++)
Console.Write(arr[i] + " ");
}
}
// JavaScript Program to left rotate the array by d positions
// using Juggling Algorithm
// Function to rotate array
function rotateArr(arr, d) {
let n = arr.length;
// Handle the case where d > size of array
d %= n;
// Calculate the number of cycles in the rotation
let cycles = gcd(n, d);
// Process each cycle
for (let i = 0; i < cycles; i++) {
// Start element of current cycle
let startEle = arr[i];
// Start index of current cycle
let currIdx = i, nextIdx;
// Rotate elements till we reach the start of cycle
while (true) {
nextIdx = (currIdx + d) % n;
if (nextIdx === i)
break;
// Update the next index with the current element
arr[currIdx] = arr[nextIdx];
currIdx = nextIdx;
}
// Copy the start element of current cycle at the last
// index of the cycle
arr[currIdx] = startEle;
}
}
// Helper function to compute GCD
function gcd(a, b) {
while (b !== 0) {
let temp = b;
b = a % b;
a = temp;
}
return a;
}
let arr = [1, 2, 3, 4, 5, 6];
let d = 2;
rotateArr(arr, d);
// Print the rotated array
console.log(arr.join(" "));
Time Complexity: O(n), as all the cycles are independent and we are not visiting any element more than once.
Auxiliary Space: O(1)
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