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 

C++

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// 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;
}

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Java

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// 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.

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Python3

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# 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 prcurrent 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 prnot 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

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C#

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// 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 */

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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.

Related Article:
Find four elements a, b, c and d in an array such that a+b = c+d

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