Find pair with greatest product in array

Given an array of n elements, the task is to find the greatest number such that it is product of two elements of given array. If no such element exists, print -1. Elements are within the range of 1 to 10^5.

Examples :

Input :  arr[] = {10, 3, 5, 30, 35}
Output:  30
Explanation: 30 is the product of 10 and 3.

Input :  arr[] = {2, 5, 7, 8}
Output:  -1
Explanation: Since, no such element exists.

Input :  arr[] = {10, 2, 4, 30, 35}
Output:  -1

Input :  arr[] = {10, 2, 2, 4, 30, 35}
Output:  4

Input  : arr[] = {17, 2, 1, 35, 30}
Output : 35

A naive approach is to pick an element and then check for each pair product if equal to that number and update the max if the number is maximum,repeat until whole array gets traversed takes O(n^3) time.

C++

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// C++ program to find a pair with product
// in given array.
#include<bits/stdc++.h>
using namespace std;
  
// Function to find greatest number that us
int findGreatest( int arr[] , int n)
{
    int result = -1;
    for (int i = 0; i < n ; i++)
        for (int j = 0; j < n-1; j++)
            for (int k = j+1 ; k < n  ; k++)
                if (arr[j] * arr[k] == arr[i])
                    result = max(result, arr[i]);
    return result;
}
  
// Driver code
int main()
{
    // Your C++ Code
    int arr[] = {30, 10, 9, 3, 35};
    int n = sizeof(arr)/sizeof(arr[0]);
  
    cout << findGreatest(arr, n);
  
    return 0;
}

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Java

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// Java program to find a pair 
// with product in given array.
import java.io.*;
  
class GFG{
  
static int findGreatest( int []arr , int n)
{
    int result = -1;
    for (int i = 0; i < n ; i++)
        for (int j = 0; j < n-1; j++)
            for (int k = j+1 ; k < n ; k++)
                if (arr[j] * arr[k] == arr[i])
                    result = Math.max(result, arr[i]);
    return result;
}
  
    // Driver code
    static public void main (String[] args)
    {
        int []arr = {30, 10, 9, 3, 35};
        int n = arr.length;
  
        System.out.println(findGreatest(arr, n));
    }
}
  
//This code is contributed by vt_m.

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Python 3

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# Python 3 program to find a pair 
# with product in given array.
  
# Function to find greatest number 
def findGreatest( arr , n):
  
    result = -1
    for i in range(n):
        for j in range(n - 1):
            for k in range(j + 1, n):
                if (arr[j] * arr[k] == arr[i]):
                    result = max(result, arr[i])
    return result
  
# Driver code
if __name__ == "__main__":
      
    arr = [ 30, 10, 9, 3, 35]
    n = len(arr)
  
    print(findGreatest(arr, n))
  
# This code is contributed by ita_c

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C#

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// C# program to find a pair with product
// in given array.
using System;
  
class GFG{
  
static int findGreatest( int []arr , int n)
{
    int result = -1;
    for (int i = 0; i < n ; i++)
        for (int j = 0; j < n-1; j++)
            for (int k = j+1 ; k < n ; k++)
                if (arr[j] * arr[k] == arr[i])
                    result = Math.Max(result, arr[i]);
    return result;
}
  
    // Driver code
    static public void Main ()
    {
       int []arr = {30, 10, 9, 3, 35};
       int n = arr.Length;
  
       Console.WriteLine(findGreatest(arr, n));
    }
}
  
//This code is contributed by vt_m.

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PHP

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<?php
// PHP program to find a pair 
// with product in given array.
  
// Function to find 
// greatest number 
function findGreatest($arr , $n)
{
    $result = -1;
    for ($i = 0; $i < $n ; $i++)
        for ($j = 0; $j < $n - 1; $j++)
            for ($k = $j + 1 ; $k < $n ; $k++)
                if ($arr[$j] * $arr[$k] == $arr[$i])
                    $result = max($result, $arr[$i]);
    return $result;
}
  
// Driver code
$arr = array(30, 10, 9, 3, 35);
$n = count($arr);
  
echo findGreatest($arr, $n);
  
// This code is contributed by anuj_67.
?>

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Output :

30

An efficient method follows below implementation:-

  1. Create an empty hash table and store all array elements in it.
  2. Sort the array in ascending order.
  3. Pick elements one by one from end of the array.
  4. And check if there exists a pair whose product is equal to that number. In this efficiency can be achieved. The idea is to reach till sqrt of that number. If we don’t get the pair till sqrt that means no such pair exists. We use hash table to make sure that we can find other element of pair in O(1) time.
  5. Repeat steps 2 to 3 until we get the element or whole array gets traversed.

Below is the implementation.

C++

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// C++ program to find the largest product number
#include<bits/stdc++.h>
using namespace std;
  
// Function to find greatest number
int findGreatest(int arr[], int n)
{
    // Store occurrences of all elements in hash
    // array
    unordered_map<int, int> m;
    for (int i = 0 ; i < n; i++)
        m[arr[i]]++;
  
    // Sort the array and traverse all elements from
    // end.
    sort(arr, arr+n);
  
    for (int i=n-1; i>1; i--)
    {
        // For every element, check if there is another
        // element which divides it.
        for (int j=0; j<i && arr[j]<=sqrt(arr[i]); j++)
        {
            if (arr[i] % arr[j] == 0)
            {
                int result = arr[i]/arr[j];
  
                // Check if the result value exists in array
                // or not if yes the return arr[i]
                if (result != arr[j] && m[result] > 0)
                    return arr[i];
  
                // To handle the case like arr[i] = 4 and
                // arr[j] = 2
                else if (result == arr[j] && m[result] > 1)
                    return arr[i];
            }
        }
    }
    return -1;
}
  
// Drivers code
int main()
{
    int arr[] = {17, 2, 1, 15, 30};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << findGreatest(arr, n);
    return 0;
}

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Java

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// Java program to find the largest product number
import java.util.*;
  
class GFG 
{
  
    // Function to find greatest number
    static int findGreatest(int arr[], int n) 
    {
        // Store occurrences of all 
        // elements in hash array
        Map<Integer, Integer> m = new HashMap<>();
        for (int i = 0; i < n; i++) 
        {
            if (m.containsKey(arr[i])) 
            {
                m.put(arr[i], m.get(arr[i]) + 1);
            
            else 
            {
                m.put(arr[i], m.get(arr[i]));
            }
        }
  
        // m[arr[i]]++;
        // Sort the array and traverse 
        // all elements from end.
        Arrays.sort(arr);
  
        for (int i = n - 1; i > 1; i--) 
        {
            // For every element, check if there is another
            // element which divides it.
            for (int j = 0; j < i && 
                arr[j] <= Math.sqrt(arr[i]); j++) 
            {
                if (arr[i] % arr[j] == 0
                {
                    int result = arr[i] / arr[j];
  
                    // Check if the result value exists in array
                    // or not if yes the return arr[i]
                    if (result != arr[j] && 
                        m.get(result) == null|| m.get(result) > 0)
                    {
                        return arr[i];
                    
                      
                    // To handle the case like arr[i] = 4
                    // and arr[j] = 2
                    else if (result == arr[j] && m.get(result) > 1
                    {
                        return arr[i];
                    }
                }
            }
        }
        return -1;
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        int arr[] = {17, 2, 1, 15, 30};
        int n = arr.length;
        System.out.println(findGreatest(arr, n));
    }
  
// This code is contributed by PrinciRaj1992

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Python

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# Python3 program to find the largest product number
from math import sqrt
  
# Function to find greatest number
def findGreatest(arr, n):
  
    # Store occurrences of all elements in hash
    # array
    m = dict()
  
    for i in arr:
        m[i] = m.get(i, 0) + 1
  
    # Sort the array and traverse all elements from
    # end.
    arr=sorted(arr)
  
    for i in range(n - 1, 0, -1):
          
        # For every element, check if there is another
        # element which divides it.
        j = 0
        while(j < i and arr[j] <= sqrt(arr[i])):
  
            if (arr[i] % arr[j] == 0):
  
                result = arr[i]//arr[j]
  
                # Check if the result value exists in array
                # or not if yes the return arr[i]
                if (result != arr[j] and (result in m.keys() )and m[result] > 0):
                    return arr[i]
  
                # To handle the case like arr[i] = 4 and
                # arr[j] = 2
                elif (result == arr[j] and (result in m.keys()) and m[result] > 1):
                    return arr[i]
  
            j += 1
  
  
    return -1
  
# Drivers code
arr= [17, 2, 1, 15, 30]
n = len(arr)
print(findGreatest(arr, n))
  
# This code is contributed by mohit kumar

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C#

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// C# program to find the largest product number
using System;
using System.Collections.Generic;
  
class GFG 
{
  
    // Function to find greatest number
    static int findGreatest(int []arr, int n) 
    {
        // Store occurrences of all 
        // elements in hash array
        Dictionary<int,int> m = new Dictionary<int,int> ();
        for (int i = 0; i < n; i++) 
        {
            if (m.ContainsKey(arr[i])) 
            {
                var a = m[arr[i]] + 1;
                  
                // m.Remove(arr[i]);
                m.Add(arr[i], a);
            
            else
            {
                m.Add(arr[i], arr[i]);
            }
        }
  
        // m[arr[i]]++;
        // Sort the array and traverse 
        // all elements from end.
        Array.Sort(arr);
  
        for (int i = n - 1; i > 1; i--) 
        {
            // For every element, check if there is another
            // element which divides it.
            for (int j = 0; j < i && 
                arr[j] <= Math.Sqrt(arr[i]); j++) 
            {
                if (arr[i] % arr[j] == 0) 
                {
                    int result = arr[i] / arr[j];
  
                    // Check if the result value exists in array
                    // or not if yes the return arr[i]
                    if (result != arr[j] && 
                        m[result] == null|| m[result] > 0)
                    {
                        return arr[i];
                    
                      
                    // To handle the case like arr[i] = 4
                    // and arr[j] = 2
                    else if (result == arr[j] && m[result] > 1) 
                    {
                        return arr[i];
                    }
                }
            }
        }
        return -1;
    }
  
    // Driver code
    public static void Main(String[] args) 
    {
        int []arr = {17, 2, 1, 15, 30};
        int n = arr.Length;
        Console.WriteLine(findGreatest(arr, n));
    }
  
// This code contributed by Rajput-Ji

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Output :

30

Time Complexity : O(nlogn)

This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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