# Page Faults in LFU | Implementation

• Difficulty Level : Hard
• Last Updated : 02 Jun, 2021

Overview :

In this, it is using the concept of paging for memory management, a page replacement algorithm is needed to decide which page needs to be replaced when the new page comes in. Whenever a new page is referred to and is not present in memory, the page fault occurs and the Operating System replaces one of the existing pages with a newly needed page. LFU is one such page replacement policy in which the least frequently used pages are replaced. If the frequency of pages is the same, then the page that has arrived first is replaced first.

Example :

Given a sequence of pages in an array of pages[] of length N and memory capacity C, find the number of page faults using the Least Frequently Used (LFU) Algorithm.

Example-1 :

```Input : N = 7, C = 3
pages = { 1, 2, 3, 4, 2, 1, 5 }
Output :
Page Faults = 6
Page Hits = 1```

Explanation :

Capacity is 3, thus, we can store maximum 3 pages at a time.

Page 1 is required, since it is not present,

it is a page fault : page fault = 1

Page 2 is required, since it is not present,

it is a page fault : page fault = 1 + 1 = 2

Page 3 is required, since it is not present,

it is a page fault : page fault = 2 + 1 = 3

Page 4 is required, since it is not present,

it replaces LFU page 1 : page fault = 3 + 1 = 4

Page 2 is required, since it is present,

it is not a page fault : page fault = 4 + 0 = 4

Page 1 is required, since it is not present,

it replaces LFU page 3 : page fault = 4 + 1 = 5

Page 5 is required, since it is not present,

it replaces LFU page 4 : page fault = 5 + 1 = 6

Example-2 :

```Input : N = 9, C = 4
pages = { 5, 0, 1, 3, 2, 4, 1, 0, 5 }
Output :
Page Faults = 8
Page Hits = 1```

Explanation :

Capacity is 4, thus, we can store maximum 4 pages at a time.

Page 5 is required, since it is not present,

it is a page fault : page fault = 1

Page 0 is required, since it is not present,

it is a page fault : page fault = 1 + 1 = 2

Page 1 is required, since it is not present,

it is a page fault : page fault = 2 + 1 = 3

Page 3 is required, since it is not present,

it is a page fault : page fault = 3 + 1 = 4

Page 2 is required, since it is not present,

it replaces LFU page 5 : page fault = 4 + 1 = 5

Page 4 is required, since it is not present,

it replaces LFU page 0 : page fault = 5 + 1 = 6

Page 1 is required, since it is present,

it is not a page fault : page fault = 6 + 0 = 6

Page 0 is required, since it is not present,

it replaces LRU page 3 : page fault = 6 + 1 = 7

Page 5 is required, since it is not present,

it replaces LFU page 2 : page fault = 7 + 1 = 8

Algorithm :

```Step-1 : Initialize count as 0.
Step-2 : Create a vector / array of size equal to memory capacity.
Create a map to store frequency of pages
Step-3 : Traverse elements of pages[]
Step-4 : In each traversal:
if(element is present in memory):
remove the element and push the element at the end
increase its frequency
else:
if(memory is full)
remove the first element and decrease frequency of 1st element
Increment count
push the element at the end and increase its frequency
Compare frequency with other pages starting from the 2nd last page
Sort the pages based on their frequency and time at which they arrive
if frequency is same, then, the page arriving first must be placed first       ```

Following is the implementation of the above algorithm.

## C++

 `// C++ program to illustrate``// page faults in LFU`` ` `#include ``using` `namespace` `std;`` ` `/* Counts no. of page faults */``int` `pageFaults(``int` `n, ``int` `c, ``int` `pages[])``{``    ``// Initialise count to 0``    ``int` `count = 0;`` ` `    ``// To store elements in memory of size c``    ``vector<``int``> v;``    ``// To store frequency of pages``    ``unordered_map<``int``, ``int``> mp;`` ` `    ``int` `i;``    ``for` `(i = 0; i <= n - 1; i++) {`` ` `        ``// Find if element is present in memory or not``        ``auto` `it = find(v.begin(), v.end(), pages[i]);`` ` `        ``// If element is not present``        ``if` `(it == v.end()) {`` ` `            ``// If memory is full``            ``if` `(v.size() == c) {`` ` `                ``// Decrease the frequency``                ``mp[v]--;`` ` `                ``// Remove the first element as``                ``// It is least frequently used``                ``v.erase(v.begin());``            ``}`` ` `            ``// Add the element at the end of memory``            ``v.push_back(pages[i]);``            ``// Increase its frequency``            ``mp[pages[i]]++;`` ` `            ``// Increment the count``            ``count++;``        ``}``        ``else` `{`` ` `            ``// If element is present``            ``// Remove the element``            ``// And add it at the end``            ``// Increase its frequency``            ``mp[pages[i]]++;``            ``v.erase(it);``            ``v.push_back(pages[i]);``        ``}`` ` `        ``// Compare frequency with other pages``        ``// starting from the 2nd last page                 ``        ``int` `k = v.size() - 2;`` ` `        ``// Sort the pages based on their frequency ``        ``// And time at which they arrive``        ``// if frequency is same``        ``// then, the page arriving first must be placed first``        ``while` `(mp[v[k]] > mp[v[k + 1]] && k > -1) {``            ``swap(v[k + 1], v[k]);``            ``k--;``        ``}``    ``}`` ` `    ``// Return total page faults``    ``return` `count;``}`` ` `/* Driver program to test pageFaults function*/``int` `main()``{`` ` `    ``int` `pages[] = { 1, 2, 3, 4, 2, 1, 5 };``    ``int` `n = 7, c = 3;`` ` `    ``cout << ``"Page Faults = "` `<< pageFaults(n, c, pages)``         ``<< endl;``    ``cout << ``"Page Hits = "` `<< n - pageFaults(n, c, pages);``    ``return` `0;``}`` ` `// This code is contributed by rajsanghavi9.`

Output

```Page Faults = 6
Page Hits = 1```

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