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.

Example:

Input: arr[] = {{11, 20}, {30, 40}, {5, 10}, {40, 30}, {10, 5}} Output: Following pairs have symmetric pairs (30, 40) (5, 10)

**We strongly recommend you to minimize your browser and try this yourself first.**

A **Simple Solution** is to go through every pair, and check every other pair for symmetric. This solution requires O(n^{2}) 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).

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.

Following are C++ and Java implementation of this idea.

## C/C++

#include<bits/stdc++.h> using namespace std; // A C++ program to find all symmetric pairs in a given array of pairs // Print all pairs that have a symmetric counterpart void findSymPairs(int arr[][2], int row) { // Creates an empty hashMap hM unordered_map<int, int> hM; // Traverse through the given array for (int i = 0; i < row; i++) { // First and second elements of current pair int first = arr[i][0]; int sec = arr[i][1]; // If found and value in hash matches with first // element of this pair, we found symmetry if (hM.find(sec) != hM.end() && hM[sec] == first) cout << "(" << sec << ", " << first << ")" <<endl; else // Else put sec element of this pair in hash hM[first] = sec; } } // Drive method int main() { int arr[5][2]; arr[0][0] = 11; arr[0][1] = 20; arr[1][0] = 30; arr[1][1] = 40; arr[2][0] = 5; arr[2][1] = 10; arr[3][0] = 40; arr[3][1] = 30; arr[4][0] = 10; arr[4][1] = 5; findSymPairs(arr, 5); } //This is contributed by Chhavi

## Java

// A Java program to find all symmetric pairs in a given array of pairs import java.util.HashMap; class SymmetricPairs { // 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); } } // Drive method public static void main(String arg[]) { int arr[][] = new int[5][2]; arr[0][0] = 11; arr[0][1] = 20; arr[1][0] = 30; arr[1][1] = 40; arr[2][0] = 5; arr[2][1] = 10; arr[3][0] = 40; arr[3][1] = 30; arr[4][0] = 10; arr[4][1] = 5; findSymPairs(arr); } }

Output:

Following pairs have symmetric pairs (30, 40) (5, 10)

Time Complexity of this solution is O(n) under the assumption that hash search and insert methods work in O(1) time.

This article is contributed by **Shivam Agrawal**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.