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Program for K Most Recently Used (MRU) Apps

  • Difficulty Level : Basic
  • Last Updated : 29 Aug, 2021

Given an integer K and an array arr[] of N integers which contains the ids of the opened apps in a system where 
 

  1. arr[0] is the app currently in use
  2. arr[1] is the app which was most recently used and
  3. arr[N – 1] is the app which was least recently used.

The task is to print the contents of the array when the user using the system presses Alt + Tab exactly K number of times. Note that after pressing Alt + Tab key, app opening pointer will move through apps from 0th index towards right, depending upon the number of presses, so the app on which the press ends will shift to 0th index, because that will become the most recently opened app.
Examples: 
 

Input: arr[] = {3, 5, 2, 4, 1}, K = 3 
Output: 4 3 5 2 1 
User want to switch to the app with id 4, it’ll become the currently active app and the previously active app (with id 3) will be the most recently used app.
Input: arr[] = {5, 7, 2, 3, 4, 1, 6}, K = 10 
Output: 3 5 7 2 4 1 6 
 

 

Approach: Get the index of the app to which the user wants to switch i.e. appIndex = K % N. Now, the current active app will be arr[appIndex] and all the other apps in the index range [0, appIndex – 1] will have to be shifted by 1 element towards the right.
Below is the implementation of the above approach: 
 



C++14




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to update the array
// in most recently used fashion
void mostRecentlyUsedApps(int* arr, int N, int K)
{
    int app_index = 0;
 
    // Finding the end index after K presses
    app_index = (K % N);
 
    // Shifting elements by 1 towards the found index
    // on which the K press ends
    int x = app_index, app_id = arr[app_index];
    while (x > 0) {
        arr[x] = arr[--x];
    }
 
    // Update the current active app
    arr[0] = app_id;
}
 
// Utility function to print
// the contents of the array
void printArray(int* arr, int N)
{
    for (int i = 0; i < N; i++)
        cout << arr[i] << " ";
}
 
// Driver code
int main()
{
    int K = 3;
    int arr[] = { 3, 5, 2, 4, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    mostRecentlyUsedApps(arr, N, K);
    printArray(arr, N);
    return 0;
}

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to update the array
// in most recently used fashion
static void mostRecentlyUsedApps(int []arr, int N, int K)
{
    int app_index = 0;
 
    // Finding the end index after K presses
    app_index = (K % N);
 
    // Shifting elements by 1 towards the found index
    // on which the K press ends
    int x = app_index, app_id = arr[app_index];
    while (x > 0)
    {
        arr[x] = arr[--x];
    }
 
    // Update the current active app
    arr[0] = app_id;
}
 
// Utility function to print
// the contents of the array
static void printArray(int []arr, int N)
{
    for (int i = 0; i < N; i++)
        Console.Write(arr[i]+" ");
}
 
// Driver code
static void Main()
{
    int K = 3;
    int []arr = { 3, 5, 2, 4, 1 };
    int N = arr.Length;
 
    mostRecentlyUsedApps(arr, N, K);
    printArray(arr, N);
}
}
 
// This code is contributed by mits

Javascript




<script>
// Javascript implementation of the approach
 
// Function to update the array
// in most recently used fashion
function mostRecentlyUsedApps(arr, N, K) {
    let app_index = 0;
 
    // Finding the end index after K presses
    app_index = (K % N);
 
    // Shifting elements by 1 towards the found index
    // on which the K press ends
    let x = app_index, app_id = arr[app_index];
    while (x > 0) {
        arr[x] = arr[--x];
    }
 
    // Update the current active app
    arr[0] = app_id;
}
 
// Utility function to print
// the contents of the array
function printArray(arr, N) {
    for (let i = 0; i < N; i++)
        document.write(arr[i] + " ");
}
 
// Driver code
 
let K = 3;
let arr = [3, 5, 2, 4, 1];
let N = arr.length;
 
mostRecentlyUsedApps(arr, N, K);
printArray(arr, N);
 
// This code is contributed by gfgking.
</script>

Java




/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
    public static void main(String[] args)
    {
        int[] num = { 3, 5, 2, 4, 1 };
        int size = num.length;
        // 8,6,7,9,0,2,1,12,89
        int d = 3;
 
        int appIndex = d % size;
        int appId = num[appIndex];
 
        for (int i = appIndex; i > 0; i--) {
            num[i] = num[i - 1];
        }
        num[0] = appId;
 
        for (int i = 0; i < num.length; i++) {
            System.out.print(" " + num[i]);
        }
    }
}
Output
4 3 5 2 1 

Time Complexity: O(N)
 

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