Given an array of pairs, find all symmetric pairs in it - GeeksforGeeks

Given an array of pairs, find all symmetric pairs in it - GeeksforGeeks
Two pairs (a, b) and (c, d) are said to be symmetric if c is equal to b and a is equal to d. For example (10, 20) and (20, 10) are symmetric. Given an array of pairs find all symmetric pairs in it.
It may be assumed that first elements of all pairs are distinct.

An Efficient Solution is to use Hashing. First element of pair is used as key and second element is used as value. The idea is traverse all pairs one by one. For every pair, check if its second element is in hash table. If yes, then compare the first element with value of matched entry of hash table. If the value and the first element match, then we found symmetric pairs. Else, insert first element as key and second element as value.

    // Print all pairs that have a symmetric counterpart

    static void findSymPairs(int arr[][])

    {

        // Creates an empty hashMap hM

        HashMap<Integer, Integer> hM = new HashMap<Integer, Integer>();

  

        // Traverse through the given array

        for (int i = 0; i < arr.length; i++)

        {

            // First and second elements of current pair

            int first = arr[i][0];

            int sec   = arr[i][1];

             

            // Look for second element of this pair in hash

            Integer val = hM.get(sec);

  

            // If found and value in hash matches with first

            // element of this pair, we found symmetry

            if (val != null && val == first)

               System.out.println("(" + sec + ", " + first + ")");

                

            else  // Else put sec element of this pair in hash

               hM.put(first, sec);

        }

    }

A Simple Solution is to go through every pair, and check every other pair for symmetric. This solution requires O(n2) time.

A Better Solution is to use sorting. Sort all pairs by first element. For every pair, do binary search for second element in the given array, i.e., check if second element of this pair exists as first element in array. If found, then compare first element of pair with second element. Time Complexity of this solution is O(nLogn).

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