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

08 Jul, 2022
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++
Java
Python3
C#
Javascript
// 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 = cdvoid 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 codeint main(){    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8};    int n = sizeof(arr)/sizeof(int);    findPairs(arr, n);    return 0;}
// Java program to find four elements a, b, c// and d in array such that ab = cdimport 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 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 = cddef 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 codearr = [1, 2, 3, 4, 5, 6, 7, 8]n = len(arr)findPairs(arr, n) # This code is contributed# by mohit kumar
// C# program to find four elements a, b, c// and d in array such that ab = cdusing 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 */
<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>
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.
Related Article: Find four elements a, b, c and d in an array such that a+b = c+d
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