Find pair with greatest product in array
Last Updated :
23 Jul, 2025
Given an array of n elements, the task is to find the greatest number such that it is the product of two elements of the 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 is equal to that number and update the max if the number is maximum, repeat until the whole array gets traversed takes O(n^3) time.
C++
// 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[] = {10, 3, 5, 30, 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 = {10, 3, 5, 30, 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 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 = [10, 3, 5, 30, 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 = {10, 3, 5, 30, 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 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(10, 3, 5, 30, 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 = [10, 3, 5, 30, 35];
let n = arr.length;
document.write(findGreatest(arr, n));
// This code is contributed by splevel62.
</script>
Time Complexity: O(n3)
Auxiliary Space: O(1)
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.
Below is the implementation.
C++
// 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] && 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 code
int main()
{
int arr[] = {10, 3, 5, 30, 35};
int n = sizeof(arr) / sizeof(arr[0]);
cout << findGreatest(arr, n);
return 0;
}
// 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)
{
Arrays.sort(arr);
HashMap<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < n; i++)
map.put(arr[i],
map.getOrDefault(arr[i], 0) + 1);
for (int i = n - 1; i > 1; i--) {
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];
if (result != arr[j]
&& map.containsKey(result)) {
if (result == arr[i]) {
if (map.get(arr[i]) > 1)
return arr[i];
}
else
return arr[i];
}
else if (result == arr[j]
&& map.get(result) > 1)
return arr[i];
}
}
}
return -1;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {10, 3, 5, 30, 35};
int n = arr.length;
System.out.println(findGreatest(arr, n));
}
}
// This code is contributed by Krishnendu Ghosh
# 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 = [10, 3, 5, 30, 35]
n = len(arr)
print(findGreatest(arr, n))
# This code is contributed by mohit kumar
// 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 = {10, 3, 5, 30, 35 };
int n = arr.Length;
Console.WriteLine(findGreatest(arr, n));
}
}
// This code is contributed by vt_m.
<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=[10, 3, 5, 30, 35];
let n = arr.length;
document.write(findGreatest(arr, n));
// This code is contributed by avanitrachhadiya2155
</script>
Time Complexity: O(n log n)
Auxiliary Space: O(n)
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