Question

for java:

NOTE: Make sure, you have exception handling code in place for each exceptional condition for every question. Also, make sure you have tested all the codes with various boundary conditions.

Given a list of pairs; Give an efficient method (O(n)) to print all “Symmetric Pairs”. A pair is symmetric if both pair (i, j) and pair (j, i) exist in the list. For example, in { {3, 1}, {2, 6}, {3, 5}, {7, 4}, {5, 3}, {8, 7} } , note that {3, 5} is already present in the list when you encounter {5, 3}, that means they are symmetric pair, so print this pair when you encounter {5, 3}. [Hint: use hash table]
Algorithm:
1) Read the pairs of elements one by one and insert them into a hash table. For each pair, consider the first element as a key and second element as value.
2) While inserting the key elements check if the hashing of the second element of the current pair is the same as the first number of the current pair.
3) If they are the same, this means that the pairs are symmetric and output that pair.
4) Otherwise, insert that element into that. That means, use the first number of the pair as key and the second number as value and insert them into the hash table.
5) By the time we complete the scanning of all pairs, we have output all the symmetric pairs.

Answer 1

import java.util.*;
public class Main {
public static void main(String[] args)
{
Hashtable<Integer, Integer> t = new Hashtable<Integer, Integer>();
Scanner sc=new Scanner(System.in);
int n=Integer.parseInt(sc.nextLine());
int a,b;
System.out.println("Enter pair of values with space");
String s="";
for(int i=0;i<n;i++)
{
s=sc.nextLine();
String d[]=s.split(" ");
a=Integer.parseInt(d[0]);
b=Integer.parseInt(d[1]);
Integer val = t.get(b);
if (val != null && val == a)
System.out.println("(" + a + ", " + b + ")");
else
t.put(a, b);
}
}
}