Find pair with greatest product in array - GeeksforGeeks

28 Aug, 2021
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++
Java
Python 3
C#
PHP
Javascript
// C++ program to find a pair with product// in given array.#include<bits/stdc++.h>using namespace std; // Function to find greatest number that usint 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 codeint main(){    // Your C++ Code    int arr[] = {30, 10, 9, 3, 35};    int n = sizeof(arr)/sizeof(arr[0]);     cout << findGreatest(arr, n);     return 0;}
// 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.
# Python 3 program to find a pair# with product in given array. # Function to find greatest numberdef 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 codeif __name__ == "__main__":         arr = [ 30, 10, 9, 3, 35]    n = len(arr)     print(findGreatest(arr, n)) # This code is contributed by ita_c
// 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.
<?php// PHP program to find a pair// with product in given array. // Function to find// greatest numberfunction 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.?>
<script> // Javascript program to find a pair// with product in given array. function findGreatest(arr , n){    let result = -1;    for (let i = 0; i < n ; i++)        for (let j = 0; j < n-1; j++)            for (let k = j+1 ; k < n ; k++)                if (arr[j] * arr[k] == arr[i])                    result = Math.max(result, arr[i]);    return result;}      // Driver code        let arr = [30, 10, 9, 3, 35];    let n = arr.length;       document.write(findGreatest(arr, n));             // This code is contributed by splevel62.       </script>
Output :  
30
An efficient method follows below implementation:-  
Create an empty hash table and store all array elements in it.Sort the array in ascending order.Pick elements one by one from end of the array.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.Repeat steps 2 to 3 until we get the element or whole array gets traversed.
Create an empty hash table and store all array elements in it.
Sort the array in ascending order.
Pick elements one by one from end of the array.
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.
Repeat steps 2 to 3 until we get the element or whole array gets traversed.
Below is the implementation. 
C++
Java
Python
C#
Javascript
// C++ program to find the largest product number#include <bits/stdc++.h>using namespace std; // Function to find greatest numberint 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] && result!=arr[i] && 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 codeint main(){    int arr[] = { 17, 2, 1, 15, 30 };    int n = sizeof(arr) / sizeof(arr[0]);    cout << findGreatest(arr, n);    return 0;}
// Java program to find the largest product numberimport 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
# Python3 program to find the largest product numberfrom math import sqrt # Function to find greatest numberdef 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 codearr= [17, 2, 1, 15, 30]n = len(arr)print(findGreatest(arr, n)) # This code is contributed by mohit kumar
// C# program to find the largest product numberusing 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
<script> // Javascript program to find the largest product number         // Function to find greatest number    function findGreatest(arr,n)    {        // Store occurrences of all        // elements in hash array        let m = new Map();        for (let i = 0; i < n; i++)        {            if (m.has(arr[i]))            {                m.set(arr[i], m[arr[i]] + 1);            }            else            {                m.set(arr[i], m.get(arr[i]));            }        }                 // m[arr[i]]++;        // Sort the array and traverse        // all elements from end.        arr.sort(function(a,b){return a-b;});                 for (let i = n - 1; i > 1; i--)        {            // For every element, check if there is another            // element which divides it.            for (let j = 0; j < i &&                arr[j] <= Math.sqrt(arr[i]); j++)            {                if (arr[i] % arr[j] == 0)                {                    let result = Math.floor(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    let arr=[17, 2, 1, 15, 30];    let n = arr.length;    document.write(findGreatest(arr, n));         // This code is contributed by avanitrachhadiya2155     </script>
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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
vt_m
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