在数组中找到具有最佳产品的对

给定一个n个元素的数组，任务是找到最大数量，使其是给定数组的两个元素的乘积。如果不存在这样的元素，则打印-1。元素在1到10 ^ 5的范围内。
例子 ：

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

一个简单的方法是选择一个元素，然后检查每个对乘积是否等于该数量，如果数量最大则更新最大值，重复直到遍历整个数组需要O(n ^ 3)的时间。

C++

// C++ program to find a pair with product
// in given array.
#include
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;
}

Java

// 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

# 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

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.

Javascript

 
// 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.
       

---

输出 ：

30

以下是一种有效的方法：-

1. 创建一个空的哈希表，并将所有数组元素存储在其中。
2. 以升序对数组进行排序。
3. 从数组末尾一个接一个地选择元素。
4. 并检查是否存在一对乘积等于该对的乘积。这样可以达到效率。这个想法是要达到该数目的平方根。如果直到sqrt才得到该对，则表示不存在这样的对。我们使用哈希表来确保可以在O(1)的时间内找到该对的其他元素。
5. 重复步骤2到3，直到遍历元素或整个数组为止。

下面是实现。

C++

// C++ program to find the largest product number
#include
using namespace std;
 
// Function to find greatest number
int findGreatest(int arr[], int n)
{
    // Store occurrences of all elements in hash
    // array
    unordered_map 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 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;
}

Java

// 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 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

Python

# 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

C＃

// 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 m = new Dictionary ();
        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

Java脚本

 
// 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
     

输出 ：

30