Program for Least Recently Used (LRU) Page Replacement algorithm

Prerequisite: Page Replacement Algorithms
In operating systems that use paging for memory management, page replacement algorithm are needed to decide which page needed to be replaced when new page comes in. Whenever a new page is referred and not present in memory, page fault occurs and Operating System replaces one of the existing pages with newly needed page. Different page replacement algorithms suggest different ways to decide which page to replace. The target for all algorithms is to reduce number of page faults.
In Least Recently Used (LRU) algorithm is a Greedy algorithm where the page to be replaced is least recently used. The idea is based on locality of reference, the least recently used page is not likely 
Let say the page reference string 7 0 1 2 0 3 0 4 2 3 0 3 2 . Initially we have 4 page slots empty. 
Initially all slots are empty, so when 7 0 1 2 are allocated to the empty slots —> 4 Page faults
0 is already there so —> 0 Page fault.
when 3 came it will take the place of 7 because it is least recently used —>1 Page fault
0 is already in memory so —> 0 Page fault. 
4 will takes place of 1 —> 1 Page Fault
Now for the further page reference string —> 0 Page fault because they are already available in the memory.

Given memory capacity (as number of pages it can hold) and a string representing pages to be referred, write a function to find number of page faults.

Let capacity be the number of pages that
memory can hold.  Let set be the current
set of pages in memory.

1- Start traversing the pages.
 i) If set holds less pages than capacity.
   a) Insert page into the set one by one until 
      the size  of set reaches capacity or all
      page requests are processed.
   b) Simultaneously maintain the recent occurred
      index of each page in a map called indexes.
   c) Increment page fault
 ii) Else 
   If current page is present in set, do nothing.
   Else 
     a) Find the page in the set that was least 
     recently used. We find it using index array.
     We basically need to replace the page with
     minimum index.
     b) Replace the found page with current page.
     c) Increment page faults.
     d) Update index of current page.

2. Return page faults.

Below is implementation of above steps.

C++

//C++ implementation of above algorithm
#include<bits/stdc++.h>
using namespace std;

// Function to find page faults using indexes
int pageFaults(int pages[], int n, int capacity)
{
    // To represent set of current pages. We use
    // an unordered_set so that we quickly check
    // if a page is present in set or not
    unordered_set<int> s;

    // To store least recently used indexes
    // of pages.
    unordered_map<int, int> indexes;

    // Start from initial page
    int page_faults = 0;
    for (int i=0; i<n; i++)
    {
        // Check if the set can hold more pages
        if (s.size() < capacity)
        {
            // Insert it into set if not present
            // already which represents page fault
            if (s.find(pages[i])==s.end())
            {
                s.insert(pages[i]);

                // increment page fault
                page_faults++;
            }

            // Store the recently used index of
            // each page
            indexes[pages[i]] = i;
        }

        // If the set is full then need to perform lru
        // i.e. remove the least recently used page
        // and insert the current page
        else
        {
            // Check if current page is not already
            // present in the set
            if (s.find(pages[i]) == s.end())
            {
                // Find the least recently used pages
                // that is present in the set
                int lru = INT_MAX, val;
                for (auto it=s.begin(); it!=s.end(); it++)
                {
                    if (indexes[*it] < lru)
                    {
                        lru = indexes[*it];
                        val = *it;
                    }
                }

                // Remove the indexes page
                s.erase(val);

                // insert the current page
                s.insert(pages[i]);

                // Increment page faults
                page_faults++;
            }

            // Update the current page index
            indexes[pages[i]] = i;
        }
    }

    return page_faults;
}

// Driver code
int main()
{
    int pages[] = {7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2};
    int n = sizeof(pages)/sizeof(pages[0]);
    int capacity = 4;
    cout << pageFaults(pages, n, capacity);
    return 0;
}

Java

// Java implementation of above algorithm

import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;

class Test
{
    // Method to find page faults using indexes
    static int pageFaults(int pages[], int n, int capacity)
    {
        // To represent set of current pages. We use
        // an unordered_set so that we quickly check
        // if a page is present in set or not
        HashSet<Integer> s = new HashSet<>(capacity);

        // To store least recently used indexes
        // of pages.
        HashMap<Integer, Integer> indexes = new HashMap<>();

        // Start from initial page
        int page_faults = 0;
        for (int i=0; i<n; i++)
        {
            // Check if the set can hold more pages
            if (s.size() < capacity)
            {
                // Insert it into set if not present
                // already which represents page fault
                if (!s.contains(pages[i]))
                {
                    s.add(pages[i]);

                    // increment page fault
                    page_faults++;
                }

                // Store the recently used index of
                // each page
                indexes.put(pages[i], i);
            }

            // If the set is full then need to perform lru
            // i.e. remove the least recently used page
            // and insert the current page
            else
            {
                // Check if current page is not already
                // present in the set
                if (!s.contains(pages[i]))
                {
                    // Find the least recently used pages
                    // that is present in the set
                    int lru = Integer.MAX_VALUE, val=Integer.MIN_VALUE;

                    Iterator<Integer> itr = s.iterator();

                    while (itr.hasNext()) {
                        int temp = itr.next();
                        if (indexes.get(temp) < lru)
                        {
                            lru = indexes.get(temp);
                            val = temp;
                        }
                    }

                    // Remove the indexes page
                    s.remove(val);
                   //remove lru from hashmap
                   indexes.remove(val);
                    // insert the current page
                    s.add(pages[i]);

                    // Increment page faults
                    page_faults++;
                }

                // Update the current page index
                indexes.put(pages[i], i);
            }
        }

        return page_faults;
    }

    // Driver method
    public static void main(String args[])
    {
        int pages[] = {7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2};

        int capacity = 4;

        System.out.println(pageFaults(pages, pages.length, capacity));
    }
}
// This code is contributed by Gaurav Miglani

Python3

# Python implementation of above algorithm
def pageFaults(pages, n, capacity):

    # To represent set of current pages. We use
    # an unordered_set so that we quickly check
    # if a page is present in set or not
    s = set()

    # To store least recently used indexes
    # of pages.
    indexes = {}

    # Start from initial page
    page_faults = 0
    for i in range(n):

        # Check if the set can hold more pages
        if len(s) < capacity:

            # Insert it into set if not present
            # already which represents page fault
            if pages[i] not in s:
                s.add(pages[i])

                # increment page fault
                page_faults += 1

            # Store the recently used index of
            # each page
            indexes[pages[i]] = i

        # If the set is full then need to perform lru
        # i.e. remove the least recently used page
        # and insert the current page
        else:

            # Check if current page is not already
            # present in the set
            if pages[i] not in s:

                # Find the least recently used pages
                # that is present in the set
                lru = float('inf')
                for page in s:
                    if indexes[page] < lru:
                        lru = indexes[page]
                        val = page

                # Remove the indexes page
                s.remove(val)

                # insert the current page
                s.add(pages[i])

                # increment page fault
                page_faults += 1

            # Update the current page index
            indexes[pages[i]] = i

    return page_faults


# Driver code
pages = [7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2]
n = len(pages)
capacity = 4
print(pageFaults(pages, n, capacity))

# This code is contributed by ishankhandelwals.

C#

// C# implementation of above algorithm
using System;
using System.Collections.Generic;

class GFG
{
    // Method to find page faults
    // using indexes
    static int pageFaults(int []pages,
                   int n, int capacity)
    {
        // To represent set of current pages.
        // We use an unordered_set so that
        // we quickly check if a page is
        // present in set or not
        HashSet<int> s = new HashSet<int>(capacity);

        // To store least recently used indexes
        // of pages.
        Dictionary<int,
                   int> indexes = new Dictionary<int,
                                                 int>();

        // Start from initial page
        int page_faults = 0;
        for (int i = 0; i < n; i++)
        {
            // Check if the set can hold more pages
            if (s.Count < capacity)
            {
                // Insert it into set if not present
                // already which represents page fault
                if (!s.Contains(pages[i]))
                {
                    s.Add(pages[i]);

                    // increment page fault
                    page_faults++;
                }

                // Store the recently used index of
                // each page
                if(indexes.ContainsKey(pages[i]))
                    indexes[pages[i]] = i;
                else
                    indexes.Add(pages[i], i);
            }

            // If the set is full then need to
            // perform lru i.e. remove the least
            // recently used page and insert
            // the current page
            else
            {
                // Check if current page is not
                // already present in the set
                if (!s.Contains(pages[i]))
                {
                    // Find the least recently used pages
                    // that is present in the set
                    int lru = int.MaxValue, val = int.MinValue;

                    foreach (int itr in s)
                    {
                        int temp = itr;
                        if (indexes[temp] < lru)
                        {
                            lru = indexes[temp];
                            val = temp;
                        }
                    }

                    // Remove the indexes page
                    s.Remove(val);

                    //remove lru from hashmap
                    indexes.Remove(val);

                    // insert the current page
                    s.Add(pages[i]);

                    // Increment page faults
                    page_faults++;
                }

                // Update the current page index
                if(indexes.ContainsKey(pages[i]))
                    indexes[pages[i]] = i;
                else
                    indexes.Add(pages[i], i);
            }
        }
        return page_faults;
    }

    // Driver Code
    public static void Main(String []args)
    {
        int []pages = {7, 0, 1, 2, 0, 3,
                       0, 4, 2, 3, 0, 3, 2};

        int capacity = 4;

        Console.WriteLine(pageFaults(pages,
                          pages.Length, capacity));
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript implementation of above algorithm

// Method to find page faults using indexes
function pageFaults(pages,n,capacity)
{
    // To represent set of current pages. We use
        // an unordered_set so that we quickly check
        // if a page is present in set or not
        let s = new Set();

        // To store least recently used indexes
        // of pages.
        let indexes = new Map();

        // Start from initial page
        let page_faults = 0;
        for (let i=0; i<n; i++)
        {
            // Check if the set can hold more pages
            if (s.size < capacity)
            {
                // Insert it into set if not present
                // already which represents page fault
                if (!s.has(pages[i]))
                {
                    s.add(pages[i]);

                    // increment page fault
                    page_faults++;
                }

                // Store the recently used index of
                // each page
                indexes.set(pages[i], i);
            }

            // If the set is full then need to perform lru
            // i.e. remove the least recently used page
            // and insert the current page
            else
            {
                // Check if current page is not already
                // present in the set
                if (!s.has(pages[i]))
                {
                    // Find the least recently used pages
                    // that is present in the set
                    let lru = Number.MAX_VALUE, val=Number.MIN_VALUE;



                    for(let itr of s.values()) {
                        let temp = itr;
                        if (indexes.get(temp) < lru)
                        {
                            lru = indexes.get(temp);
                            val = temp;
                        }
                    }

                    // Remove the indexes page
                    s.delete(val);
                   //remove lru from hashmap
                   indexes.delete(val);
                    // insert the current page
                    s.add(pages[i]);

                    // Increment page faults
                    page_faults++;
                }

                // Update the current page index
                indexes.set(pages[i], i);
            }
        }

        return page_faults;
}

 // Driver method
let pages=[7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2];
let capacity = 4;
document.write(pageFaults(pages, pages.length, capacity));


// This code is contributed by rag2127

</script>

Output:

6

Complexity Analysis :

• Time Complexity  :   average time complexity of set and map operations is O(1) and the worst-case time complexity is O(n) but O(n) is the dominant term.
• Space Complexity :  O(capacity) which is a constant and depends on the size of the input array and the size of the memory buffer.

Another approach: (Without using HashMap) 

Following are the steps to solve this problem :

1. Using a deque data structure, the program implements the page replacement algorithm.
2. A predetermined number of pages are kept in memory by the algorithm, and they are replaced as new pages are requested.
3. Using an integer array to stimulate page requests, the code keeps track the number of page faults that occur throughout the simulation.
4. The deque data structure, which is built using STL in C++, is used to maintain the pages in memory.
5. The total number of page faults that occurred throughout the simulation is given as output by the code.
   

Below is the implementation of the above approach : 

C++

// C++ program for page replacement algorithms
#include <iostream>
#include<bits/stdc++.h>
using namespace std;

int main()
{
  int capacity = 4;
  int arr[] = {7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2};

  deque<int> q(capacity);
  int count=0;
  int page_faults=0;
  deque<int>::iterator itr;
  q.clear();
  for(int i:arr)
  {

    // Insert it into set if not present
    // already which represents page fault
    itr = find(q.begin(),q.end(),i);
    if(!(itr != q.end()))
    {

      ++page_faults;

      // Check if the set can hold equal pages
      if(q.size() == capacity)
      {
        q.erase(q.begin());
        q.push_back(i);
      }
      else{
        q.push_back(i);

      }
    }
    else
    {
      // Remove the indexes page
      q.erase(itr);

      // insert the current page
      q.push_back(i);
    }

  }
  cout<<page_faults;
}

// This code is contributed by Akshit Saxena

Java

// Java program for page replacement algorithms
import java.util.ArrayList;

public class LRU {

    // Driver method
    public static void main(String[] args) {
        int capacity = 4;
        int arr[] = {7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2};

        // To represent set of current pages.We use
        // an Arraylist
        ArrayList<Integer> s=new ArrayList<>(capacity);
        int count=0;
        int page_faults=0;
        for(int i:arr)
        {
            // Insert it into set if not present
            // already which represents page fault
            if(!s.contains(i))
            {

            // Check if the set can hold equal pages
            if(s.size()==capacity)
            {
                s.remove(0);
                s.add(capacity-1,i);
            }
            else
                s.add(count,i);
                // Increment page faults
                page_faults++;
                ++count;

            }
            else
            {
                // Remove the indexes page
                s.remove((Object)i);
                // insert the current page
                s.add(s.size(),i);
            }

        }
        System.out.println(page_faults);
    }
}

Python3

# Python3 program for page replacement algorithm

# Driver code
capacity = 4
processList = [ 7, 0, 1, 2, 0, 3, 0,
                4, 2, 3, 0, 3, 2]

# List of current pages in Main Memory
s = []

pageFaults = 0
# pageHits = 0

for i in processList:

    # If i is not present in currentPages list
    if i not in s:

        # Check if the list can hold equal pages
        if(len(s) == capacity):
            s.remove(s[0])
            s.append(i)

        else:
            s.append(i)

        # Increment Page faults
        pageFaults +=1

    # If page is already there in
    # currentPages i.e in Main
    else:

        # Remove previous index of current page
        s.remove(i)

        # Now append it, at last index
        s.append(i)

print("{}".format(pageFaults))

# This code is contributed by mahi_07

C#

// C# program for page replacement algorithms
using System;
using System.Collections.Generic;

class LRU
{

    // Driver method
    public static void Main(String[] args)
    {
        int capacity = 4;
        int []arr = {7, 0, 1, 2, 0, 3, 0,
                     4, 2, 3, 0, 3, 2};

        // To represent set of current pages.
        // We use an Arraylist
        List<int> s = new List<int>(capacity);
        int count = 0;
        int page_faults = 0;
        foreach(int i in arr)
        {
            // Insert it into set if not present
            // already which represents page fault
            if(!s.Contains(i))
            {

            // Check if the set can hold equal pages
            if(s.Count == capacity)
            {
                s.RemoveAt(0);
                s.Insert(capacity - 1, i);
            }
            else
                s.Insert(count, i);

                // Increment page faults
                page_faults++;
                ++count;
            }
            else
            {
                // Remove the indexes page
                s.Remove(i);

                // insert the current page
                s.Insert(s.Count, i);
            }
        }
        Console.WriteLine(page_faults);
    }
}

// This code is contributed by Rajput-Ji

Javascript

let capacity = 4;
let arr = [7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2];

let q = [];
let count = 0;
let page_faults = 0;
let itr;
q.length = 0;

for (let i of arr)
{

  // Insert it into set if not present
  // already which represents page fault
  itr = q.indexOf(i);
  if (itr == -1) {
    page_faults++;

    // Check if the set can hold equal pages
    if (q.length == capacity) {
      q.shift();
      q.push(i);
    } else {
      q.push(i);
    }
  } else {
    // Remove the indexes page
    q.splice(itr, 1);

    // insert the current page
    q.push(i);
  }
}

console.log(page_faults);

// This code is contributed by ishankhandelwals.

Output:

6

Complexity Analysis :

• Time Complexity : O(n), as it performs a constant amount of work for each page request.
• Space Complexity : O(n+4), where n is the size of the input array and 4 is the size of the memory buffer.

Note : We can also find the number of page hits. Just have to maintain a separate count. 
If the current page is already in the memory then that must be count as Page-hit.
We will discuss other Page-replacement Algorithms in further sets.
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