Hey there!, its been quite for a while, hasn’t it?. well i got something, So my Professor gave me a Problem to solve Today, i thought you might find it interesting too, my solution is in C++, yours can be in any language, tell everyone in the comment, if you find another solution.
The Problem is:
You have been given Number of open applications(or say windows), then current application position and you’ll be given a range of useful applications (application that should not be closed), and rest application should be closed. time taken to step each window is 1 second, The useful windows will be in continues form . If unuseful windows are in continues form then the time to close all of them is only 1 second.
find the time taken to close all unuseful windows
Example:
INPUT
No. of windows: 5
range of useful windows:
2
4
current position: 3
OUTPUT
time taken is : 6
Example :
// Created by Dinesh Solanki on 04-02-2019. #include <iostream> using namespace std; int start,end,time=0,windows,pos,temp; bool fromStart=false,atEnd=false,rightDone=false,leftDone=false; void goRight(int &time) { if(!atEnd) { temp=pos; while(temp++!=(end+1)) { time++; } if(!leftDone && !fromStart) time+=time; //time+=(windows-end)+(windows-end); } rightDone=true; } void goLeft(int &time) { if(!fromStart) { temp = pos; while(temp--!=start-1) time++; if(!rightDone && !atEnd) time+=time; } leftDone=true; } int main() { cout<<"\nEnter no of Open windows:"; cin>>windows; cout<<"\nEnter Start and end of range(useful windows):"; cin>>start>>end; cout<<"\nEnter current position:"; cin>>pos; if(end==windows) atEnd=true; if(start==1) fromStart=true; if(pos-start<end-pos) { goLeft(time); goRight(time); } else { goRight(time); goLeft(time); } //time+=(start-1)+(start-1); cout<<"Time taken to close: "<<time<<" Second(s)."; return 0; }
Abstract Programmers 