Find all pairs (a,b) and (c,d) in array which satisfy ab = cd

Given an array of distinct integers, the task is to find two pairs (a, b) and (c, d) such that ab = cd, where a, b, c and d are distinct elements.

Examples:

Input  : arr[] = {3, 4, 7, 1, 2, 9, 8}
Output : 4 2 and 1 8
Product of 4 and 2 is 8 and 
also product of 1 and 8 is 8 .

Input  : arr[] = {1, 6, 3, 9, 2, 10};
Output : 6 3 and 9 2

A Simple Solution is to run four loops to generate all possible quadruples of array element. For every quadruple (a, b, c, d), check if a*b = c*d. Time complexity of this solution is O(n⁴).

An Efficient Solution of this problem is to use hashing. We use product as key and pair as value in hash table. 

1. For i=0 to n-1
2.   For j=i+1 to n-1
       a) Find  prod = arr[i]*arr[j]       
       b) If prod is not available in hash then make 
            H[prod] = make_pair(i, j) // H is hash table
       c) If product is also available in hash 
          then print previous and current elements
          of array 

Implementation:

C++

// C++ program to find four elements a, b, c
// and d in array such that ab = cd
#include<bits/stdc++.h>
using namespace std;

// Function to find out four elements in array
// whose product is ab = cd
void findPairs(int arr[], int n)
{
    bool found = false;
    unordered_map<int, pair < int, int > > H;
    for (int i=0; i<n; i++)
    {
        for (int j=i+1; j<n; j++)
        {
            // If product of pair is not in hash table,
            // then store it
            int prod = arr[i]*arr[j];
            if (H.find(prod) == H.end())
                H[prod] = make_pair(i,j);

            // If product of pair is also available in
            // then print current and previous pair
            else
            {
                pair<int,int> pp = H[prod];
                cout << arr[pp.first] << " " << arr[pp.second]
                     << " and " << arr[i]<<" "<<arr[j]<<endl;
                found = true;
            }
        }
    }
    // If no pair find then print not found
    if (found == false)
        cout << "No pairs Found" << endl;
}

//Driven code
int main()
{
    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8};
    int n = sizeof(arr)/sizeof(int);
    findPairs(arr, n);
    return 0;
}

Java

// Java program to find four elements a, b, c
// and d in array such that ab = cd
import java.io.*;
import java.util.*;

class GFG {

    public static class pair {

        int first,second;

        pair(int f, int s)
        {
            first = f;
            second = s;
        }
    };

    // Function to find out four elements
    // in array whose product is ab = cd
    public static void findPairs(int arr[], int n)
    {

        boolean found = false;
        HashMap<Integer, pair> hp =
                     new HashMap<Integer, pair>();

        for(int i = 0; i < n; i++)
        {
            for(int j = i + 1; j < n; j++)
            {

                // If product of pair is not in
                // hash table, then store it
                int prod = arr[i] * arr[j];

                if(!hp.containsKey(prod))
                    hp.put(prod, new pair(i,j));

                // If product of pair is also
                // available in then print
                // current and previous pair
                else
                {
                    pair p = hp.get(prod);
                    System.out.println(arr[p.first]
                              + " " + arr[p.second]
                              + " " + "and" + " " +
                             arr[i] + " " + arr[j]);
                    found = true;
                }
            }
        }

        // If no pair find then print not found
        if(found == false)
        System.out.println("No pairs Found");
    }

    // Driver code
    public static void main (String[] args)
    {
        int arr[] = {1, 2, 3, 4, 5, 6, 7, 8};
        int n = arr.length;
        findPairs(arr, n);
    }
}

// This code is contributed by akash1295.

Python3

# Python3 program to find four elements
# a, b, c and d in array such that ab = cd

# Function to find out four elements in array
# whose product is ab = cd
def findPairs(arr, n):

    found = False
    H = dict()

    for i in range(n):

        for j in range(i + 1, n):

            # If product of pair is not in hash table,
            # then store it
            prod = arr[i] * arr[j]

            if (prod not in H.keys()):
                H[prod] = [i, j]

            # If product of pair is also available in
            # then print current and previous pair
            else:

                pp = H[prod]
                print(arr[pp[0]], arr[pp[1]],
                      "and", arr[i], arr[j])
                found = True

    # If no pair find then print not found
    if (found == False):
        print("No pairs Found")

# Driver code
arr = [1, 2, 3, 4, 5, 6, 7, 8]
n = len(arr)
findPairs(arr, n)

# This code is contributed
# by mohit kumar

C#

// C# program to find four elements a, b, c
// and d in array such that ab = cd
using System;
using System.Collections.Generic;

class GFG
{

    public class pair
    {

        public int first,second;

        public pair(int f, int s)
        {
            first = f;
            second = s;
        }
    };

    // Function to find out four elements
    // in array whose product is ab = cd
    public static void findPairs(int[] arr, int n)
    {

        bool found = false;
        Dictionary<int, pair> hp =
                    new Dictionary<int, pair>();

        for(int i = 0; i < n; i++)
        {
            for(int j = i + 1; j < n; j++)
            {

                // If product of pair is not in
                // hash table, then store it
                int prod = arr[i] * arr[j];

                if(!hp.ContainsKey(prod))
                    hp.Add(prod, new pair(i,j));

                // If product of pair is also
                // available in then print
                // current and previous pair
                else
                {
                    pair p = hp[prod];
                    Console.WriteLine(arr[p.first]
                            + " " + arr[p.second]
                            + " " + "and" + " " +
                            arr[i] + " " + arr[j]);
                    found = true;
                }
            }
        }

        // If no pair find then print not found
        if(found == false)
        Console.WriteLine("No pairs Found");
    }

    // Driver code
    public static void Main (String[] args)
    {
        int []arr = {1, 2, 3, 4, 5, 6, 7, 8};
        int n = arr.Length;
        findPairs(arr, n);
    }
}

/* This code contributed by PrinciRaj1992 */

Javascript

<script>
// Javascript program to find four elements a, b, c
// and d in array such that ab = cd

    // Function to find out four elements
    // in array whose product is ab = cd
    function findPairs(arr,n)
    {
        let found = false;
        let  hp = new Map();

        for(let i = 0; i < n; i++)
        {
            for(let j = i + 1; j < n; j++)
            {

                // If product of pair is not in
                // hash table, then store it
                let prod = arr[i] * arr[j];

                if(!hp.has(prod))
                    hp.set(prod, [i,j]);

                // If product of pair is also
                // available in then print
                // current and previous pair
                else
                {
                    let p = hp.get(prod);
                    document.write(arr[p[0]]
                              + " " + arr[p[1]]
                              + " " + "and" + " " +
                             arr[i] + " " + arr[j]+"<br>");
                    found = true;
                }
            }
        }

        // If no pair find then print not found
        if(found == false)
            document.write("No pairs Found");
    }

    // Driver code
    let arr=[1, 2, 3, 4, 5, 6, 7, 8];
    let n = arr.length;
    findPairs(arr, n);


// This code is contributed by unknown2108
</script>

Output

1 6 and 2 3
1 8 and 2 4
2 6 and 3 4
3 8 and 4 6

Time Complexity : O(n²) assuming hash search and insert operations take O(1) time.

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