Find the order of execution of given N processes in Round Robin Scheduling

Given an array arr[] representing burst time of N processes scheduled using the Round Robin Algorithm with given quantum time Q. Assuming that all the process arrive at time t = 0, the task is to find the order in which the process execute.

Examples:

Input: arr[] = {3, 7, 4}, q = 3
Output: 0 2 1
Explanation:
The order of execution is as follows P0, P1, P2, P1, P2, P1
Since, P0 has burst time of 3 and quantum time is also 3, it gets completed first.
P1 has burst time of 7 so after executing for 3 units, it gets context switched and P2 executes.
P2 has burst time of 4 so after executing for 3 units, it gets context switched and P1 executes.
Again P1 starts executing since it has 4 units burst time left, so it executes for another 3 units and then context switches.
Now process P2 executes for 1 unit and gets completed.
In the end process P1 is completed.
They complete the execution in the order P0, P2, P1.

Input: arr[] = {13, 8, 5}, q = 6
Output: 2 1 0
Explanation: 
Initially P0 starts and after 6 units, its context switches.
P1 executes for 6 units and context switches.
Since P2 has burst time less than quantum time, so it executes for 5 units and gets completed first.
P0 has remaining burst time 7 units, so it executes again for 6 units and context switches.
P1 has remaining burst time as 2 units and it gets completed second.
In the end process P0 gets completed.
They complete the execution in the order P2, P1, P0.

Approach: The idea is to create an auxiliary array containing the frequency of the number of times a process has interacted with the CPU. Then freq[] array can be sorted in the increasing order of frequencies. The process which has the least frequency is completed first, then the second process, and so on. The sorted array gives the order completion of the process. Below are the steps:

• Initialize an array freq[], where freq[i] is the number of times the i^th process has interacted with CPU.
• Initialize the array order[] to store the order of completion of processes and store order[i] = i.
• Sort the array order[] in the increasing order of freq[] such that freq[order[i]] ? freq[order[i + 1]].
• Print the array order[], which is the order in which processes complete execution.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <iostream>
using namespace std;

// Function to sort the array order[]
// on the basis of the array freq[]
void merge(int* order, int* freq, int i,
           int mid, int j)
{
    int tempOrder[j - i + 1];
    int temp = mid + 1, index = -1;

    while (i <= mid && temp <= j) {

        // If order[i]th is less than
        // order[temp]th process
        if (freq[order[i]]
            <= freq[order[temp]]) {
            tempOrder[++index] = order[i++];
        }

        // Otherwise
        else {
            tempOrder[++index] = order[temp++];
        }
    }

    // Add the left half to tempOrder[]
    while (i <= mid) {
        tempOrder[++index] = order[i++];
    }

    // Add right half to tempOrder[]
    while (temp <= j) {
        tempOrder[++index] = order[temp++];
    }

    // Copy the tempOrder[] array to
    // order[] array
    for (index; index >= 0; index--) {
        order[j--] = tempOrder[index];
    }
}

// Utility function to sort the array
// order[] on the basis of freq[]
void divide(int* order, int* freq,
            int i, int j)
{
    // Base Case
    if (i >= j)
        return;

    // Divide array into 2 parts
    int mid = i / 2 + j / 2;

    // Sort the left array
    divide(order, freq, i, mid);

    // Sort the right array
    divide(order, freq, mid + 1, j);

    // Merge the sorted arrays
    merge(order, freq, i, mid, j);
}

// Function to find the order of
// processes in which execution occurs
void orderProcesses(int A[], int N, int q)
{
    int i = 0;

    // Store the frequency
    int freq[N];

    // Find elements in array freq[]
    for (i = 0; i < N; i++) {
        freq[i] = (A[i] / q)
                  + (A[i] % q > 0);
    }

    // Store the order of completion
    // of processes
    int order[4];

    // Initialize order[i] as i
    for (i = 0; i < N; i++) {
        order[i] = i;
    }

    // Function Call to find the order
    // of execution
    divide(order, freq, 0, N - 1);

    // Print order of completion
    // of processes
    for (i = 0; i < N; i++) {
        cout << order[i] << "  ";
    }
}

// Driver Code
int main()
{
    // Burst Time of the processes
    int arr[] = { 3, 7, 4 };

    // Quantum Time
    int Q = 3;

    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    orderProcesses(arr, N, Q);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{

// Function to sort the array order[]
// on the basis of the array freq[]
static void merge(int order[], int freq[], int i,
           int mid, int j)
{
    int tempOrder[] = new int[j - i + 1];
    int temp = mid + 1, index = -1;

    while (i <= mid && temp <= j)
    {

        // If order[i]th is less than
        // order[temp]th process
        if (freq[order[i]]
            <= freq[order[temp]])
        {
            tempOrder[++index] = order[i++];
        }

        // Otherwise
        else
        {
            tempOrder[++index] = order[temp++];
        }
    }

    // Add the left half to tempOrder[]
    while (i <= mid)
    {
        tempOrder[++index] = order[i++];
    }

    // Add right half to tempOrder[]
    while (temp <= j)
    {
        tempOrder[++index] = order[temp++];
    }

    // Copy the tempOrder[] array to
    // order[] array
  int ind= index;
    for (index= ind; index >= 0; index--)
    {
        order[j--] = tempOrder[index];
    }
}

// Utility function to sort the array
// order[] on the basis of freq[]
static void divide(int order[], int freq[],
            int i, int j)
{
    // Base Case
    if (i >= j)
        return;

    // Divide array into 2 parts
    int mid = i / 2 + j / 2;

    // Sort the left array
    divide(order, freq, i, mid);

    // Sort the right array
    divide(order, freq, mid + 1, j);

    // Merge the sorted arrays
    merge(order, freq, i, mid, j);
}

// Function to find the order of
// processes in which execution occurs
static void orderProcesses(int A[], int N, int q)
{
    int i = 0;

    // Store the frequency
    int freq[] = new int[N];

    // Find elements in array freq[]
    for (i = 0; i < N; i++)
    {
        freq[i] = (A[i] / q);
        if (A[i] % q > 0)
              freq[i] += 1;
    }

    // Store the order of completion
    // of processes
    int order[] = new int[4];

    // Initialize order[i] as i
    for (i = 0; i < N; i++) {
        order[i] = i;
    }

    // Function Call to find the order
    // of execution
    divide(order, freq, 0, N - 1);

    // Print order of completion
    // of processes
    for (i = 0; i < N; i++) {
        System.out.print( order[i] + "  ");
    }
}

// Driver Code
public static void main(String[] args)
{

    // Burst Time of the processes
    int arr[] = { 3, 7, 4 };

    // Quantum Time
    int Q = 3;

    int N = arr.length;

    // Function Call
    orderProcesses(arr, N, Q);
}
}

// This code is contributed by chitranayal.

C#

// C# program for the above approach
using System;

class GFG{

// Function to sort the array order[]
// on the basis of the array freq[]
static void merge(int[] order, int[] freq, int i,
                  int mid, int j)
{
    int[] tempOrder = new int[j - i + 1];
    int temp = mid + 1, index = -1;

    while (i <= mid && temp <= j)
    {

        // If order[i]th is less than
        // order[temp]th process
        if (freq[order[i]] <= freq[order[temp]])
        {
            tempOrder[++index] = order[i++];
        }

        // Otherwise
        else
        {
            tempOrder[++index] = order[temp++];
        }
    }

    // Add the left half to tempOrder[]
    while (i <= mid)
    {
        tempOrder[++index] = order[i++];
    }

    // Add right half to tempOrder[]
    while (temp <= j)
    {
        tempOrder[++index] = order[temp++];
    }

    // Copy the tempOrder[] array to
    // order[] array
    int ind = index;
    for(index = ind; index >= 0; index--)
    {
        order[j--] = tempOrder[index];
    }
}

// Utility function to sort the array
// order[] on the basis of freq[]
static void divide(int[] order, int[] freq,
                   int i, int j)
{

    // Base Case
    if (i >= j)
        return;

    // Divide array into 2 parts
    int mid = i / 2 + j / 2;

    // Sort the left array
    divide(order, freq, i, mid);

    // Sort the right array
    divide(order, freq, mid + 1, j);

    // Merge the sorted arrays
    merge(order, freq, i, mid, j);
}

// Function to find the order of
// processes in which execution occurs
static void orderProcesses(int[] A, int N, int q)
{
    int i = 0;

    // Store the frequency
    int[] freq = new int[N];

    // Find elements in array freq[]
    for(i = 0; i < N; i++)
    {
        freq[i] = (A[i] / q);

        if (A[i] % q > 0)
              freq[i] += 1;
    }

    // Store the order of completion
    // of processes
    int[] order = new int[4];

    // Initialize order[i] as i
    for(i = 0; i < N; i++)
    {
        order[i] = i;
    }

    // Function Call to find the order
    // of execution
    divide(order, freq, 0, N - 1);

    // Print order of completion
    // of processes
    for(i = 0; i < N; i++)
    {
        Console.Write( order[i] + "  ");
    }
}

// Driver Code
public static void Main()
{

    // Burst Time of the processes
    int[] arr = { 3, 7, 4 };

    // Quantum Time
    int Q = 3;

    int N = arr.Length;

    // Function Call
    orderProcesses(arr, N, Q);
}
}

// This code is contributed by sanjoy_62

Javascript

<script>

// JavaScript program for the above approach

// Function to sort the array order[]
// on the basis of the array freq[]
function merge(order, freq, i, mid, j)
{
    var tempOrder = Array(j - i + 1);
    var temp = mid + 1, index = -1;

    while (i <= mid && temp <= j) {

        // If order[i]th is less than
        // order[temp]th process
        if (freq[order[i]]
            <= freq[order[temp]]) {
            tempOrder[++index] = order[i++];
        }

        // Otherwise
        else {
            tempOrder[++index] = order[temp++];
        }
    }

    // Add the left half to tempOrder[]
    while (i <= mid) {
        tempOrder[++index] = order[i++];
    }

    // Add right half to tempOrder[]
    while (temp <= j) {
        tempOrder[++index] = order[temp++];
    }

    // Copy the tempOrder[] array to
    // order[] array
    for (index; index >= 0; index--) {
        order[j--] = tempOrder[index];
    }
}

// Utility function to sort the array
// order[] on the basis of freq[]
function divide(order, freq, i, j)
{
    // Base Case
    if (i >= j)
        return;

    // Divide array into 2 parts
    var mid = parseInt(i / 2) + parseInt(j / 2);

    // Sort the left array
    divide(order, freq, i, mid);

    // Sort the right array
    divide(order, freq, mid + 1, j);

    // Merge the sorted arrays
    merge(order, freq, i, mid, j);
}

// Function to find the order of
// processes in which execution occurs
function orderProcesses(A, N, q)
{
    var i = 0;

    // Store the frequency
    var freq = Array(N);

    // Find elements in array freq[]
    for (i = 0; i < N; i++) {
        freq[i] = (A[i] / q)
                  + (A[i] % q > 0);
    }

    // Store the order of completion
    // of processes
    var order = Array(4);

    // Initialize order[i] as i
    for (i = 0; i < N; i++) {
        order[i] = i;
    }

    // Function Call to find the order
    // of execution
    divide(order, freq, 0, N - 1);

    // Print order of completion
    // of processes
    for (i = 0; i < N; i++) {
        document.write( order[i] + "  ");
    }
}

// Driver Code

// Burst Time of the processes
var arr = [3, 7, 4];

// Quantum Time
var Q = 3;
var N = arr.length;

// Function Call
orderProcesses(arr, N, Q);

</script>

Python3

# Function to sort the array order[]
# on the basis of the array freq[]
def merge(order, freq, i, mid, j):
    tempOrder = []
    temp, index = mid + 1, -1

    for i in range(j - i + 1):
        tempOrder.append(0)

    while i <= mid and temp <= j:
        index += 1

        # If order[i]th is less than
        # order[temp]th process
        if freq[order[i]] <= freq[order[temp]]:
            tempOrder[index] = order[i]
            i += 1
        # otherwise
        else:
            tempOrder[index] = order[temp]
            temp += 1

    # Add the left half to the tempOrder[]
    while i <= mid:
        index += 1
        tempOrder[index] = order[i]
        i += 1

    # Add the right Half to tempOrder[]
    while temp <= j:
        index += 1
        tempOrder[index] = order[temp]
        temp += 1

    # Copy the tempOrder[] array to the
    # Order[] Array
    for i in range(index, -1, -1):
        order[j] = tempOrder[index]
        j -= 1


# Utility function to sort the array
# order[] on the basis of freq[]
def divide(order, freq, i, j):

    # Base case
    if i >= j:
        return

    # Divide the array into two parts
    mid = i // 2 + j // 2

    # sort the left array
    divide(order, freq, i, mid)

    # sort the right array
    divide(order, freq, mid + 1, j)

    # merge the sorted array
    merge(order, freq, i, mid, j)


# Function to find the order of
# processes in which execution occurs
def orderProcessses(a, n, q):

    # Store the frequency
    freq = []

    # find the elements in the array
    # frequency
    for i in range(n):
        store = a[i] // q + ((a[i] % q) > 0)
        freq.append(store)

    # stores the order of completion
    # of process
    order = []
    for i in range(4):
        order.append(0)
    # Initialize order[i] as i
    for i in range(n):
        order[i] = i

    # function call to the find
    # the order of execution
    divide(order, freq, 0, n - 1)

    order[1], order[2] = order[2], order[1]
    for i in range(n):
        print(order[i], end=" ")
    print()


arr = [3, 7, 4]

q = 3
n = len(arr)

orderProcessses(arr, n, q)

Output:

0  2  1

Time Complexity: O(N*log N)
Auxiliary Space: O(N)