Left or Right Position Array

Left or Right Position Array

  January 21, 2018    

---

Problem:

Link : 
https://practice.geeksforgeeks.org/problems/left-or-right-positioned-array/0

idea to use a temp array to check whether we have used all elements to make positioned array.
Here there is only 2 possibilities that either i=index or n-index-1;
Hence look at both condiotions.


import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
 {
public static boolean leftRightposition(int arr[],int n)
{
boolean visited[]=new boolean[n];
for(int i=0;i<n;++i)
{
if(arr[i]<n)
{
if(!visited[arr[i]])
visited[arr[i]]=true;
else
visited[n-1-arr[i]]=true;
}
}
for(int i=0;i<n;++i)
if(!visited[i])
return false;
return true;
}
public static void main (String[] args)
{
Scanner ab=new Scanner(System.in);
int t=ab.nextInt();
while(t-->0)
{
    int n=ab.nextInt();
    int arr[]=new int[n];
    for(int i=0;i<n;++i)
    arr[i]=ab.nextInt();
    System.out.println(leftRightposition(arr,n)?"1":"0");
}
}
}