从两个具有最大和的数组中找到对的总和

给定两个正整数和不同整数数组。任务是从两个数组中找到总和最大的对。
注意：该对应包含两个数组中的一个元素。
例子：

Input : arr1[] = {1, 2, 3},  arr2[] = {4, 5, 6}
Output : Max Sum = 9
Pair (3, 6) has the maximum sum.

Input : arr1[] = {10, 2, 3},  arr2[] = {3, 4, 7}
Output : Max Sum = 17

朴素方法：一种简单的方法是两次运行两个嵌套循环以生成所有可能的对并找到总和最大的对。
时间复杂度：O(N*M)，其中 N 是第一个数组的长度，M 是第二个数组的长度。
有效方法：一种有效的方法是观察各个数组中最大的元素将有助于最大和对。因此，任务减少以从两个数组中找到最大元素并打印它们的总和。
下面是上述方法的实现：

C++

// C++ program to find maximum sum
// pair from two arrays
 
#include 
using namespace std;
 
// Function to find maximum sum
// pair from two arrays
int maxSumPair(int arr1[], int n1, int arr2[], int n2)
{
    int max1 = INT_MIN; // max from 1st array
    int max2 = INT_MIN; // max from 2nd array
 
    // Find max from 1st array
    for (int i = 0; i < n1; i++) {
        if (arr1[i] > max1)
            max1 = arr1[i];
    }
 
    // Find max from 2nd array
    for (int i = 0; i < n2; i++) {
        if (arr2[i] > max2)
            max2 = arr2[i];
    }
 
    // Return sum of max from both arrays
    return max1 + max2;
}
 
// Driver Code
int main()
{
 
    int arr1[] = { 10, 2, 3 };
 
    int arr2[] = { 3, 4, 7 };
 
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
 
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
 
    cout << maxSumPair(arr1, n1, arr2, n2);
 
    return 0;
}

Java

//Java program to find maximum sum
//pair from two arrays
public class AAB {
 
    //Function to find maximum sum
    //pair from two arrays
    static int maxSumPair(int arr1[], int n1, int arr2[], int n2)
    {
     int max1 = Integer.MIN_VALUE; // max from 1st array
     int max2 = Integer.MIN_VALUE; // max from 2nd array
 
     // Find max from 1st array
     for (int i = 0; i < n1; i++) {
         if (arr1[i] > max1)
             max1 = arr1[i];
     }
 
     // Find max from 2nd array
     for (int i = 0; i < n2; i++) {
         if (arr2[i] > max2)
             max2 = arr2[i];
     }
 
     // Return sum of max from both arrays
     return max1 + max2;
    }
 
    //Driver Code
    public static void main(String[] args) {
         
        int arr1[] = { 10, 2, 3 };
 
         int arr2[] = { 3, 4, 7 };
 
         int n1 = arr1.length;
 
         int n2 = arr2.length;
 
         System.out.println(maxSumPair(arr1, n1, arr2, n2));
 
    }
 
}

Python3

# Python3 program to find maximum
# sum pair from two arrays
import sys
 
# Function to find maximum sum
# pair from two arrays
def maxSumPair(arr1, n1, arr2, n2):
 
    max1 = -sys.maxsize -1 # max from 1st array
    max2 = -sys.maxsize -1 # max from 2nd array
 
    # Find max from 1st array
    for i in range(0, n1):
        if (arr1[i] > max1):
            max1 = arr1[i]
     
    # Find max from 2nd array
    for i in range(0, n2):
        if (arr2[i] > max2):
            max2 = arr2[i]
     
    # Return sum of max from
    # both arrays
    return max1 + max2
 
# Driver Code
arr1 = [ 10, 2, 3 ]
 
arr2 = [ 3, 4, 7 ]
 
n1 = len(arr1)
 
n2 = len(arr2)
 
print(maxSumPair(arr1, n1, arr2, n2))
 
# This code is contributed
# by Yatin Gupta

C#

// C# program to find maximum sum
//pair from two arrays
using System;
 
public class AAB {
  
    //Function to find maximum sum
    //pair from two arrays
    static int maxSumPair(int []arr1, int n1, int []arr2, int n2)
    {
     int max1 = int.MinValue; // max from 1st array
     int max2 = int.MinValue; // max from 2nd array
  
     // Find max from 1st array
     for (int i = 0; i < n1; i++) {
         if (arr1[i] > max1)
             max1 = arr1[i];
     }
  
     // Find max from 2nd array
     for (int i = 0; i < n2; i++) {
         if (arr2[i] > max2)
             max2 = arr2[i];
     }
  
     // Return sum of max from both arrays
     return max1 + max2;
    }
  
    //Driver Code
    public static void Main() {
          
        int []arr1 = { 10, 2, 3 };
  
         int []arr2 = { 3, 4, 7 };
  
         int n1 = arr1.Length;
  
         int n2 = arr2.Length;
  
         Console.Write(maxSumPair(arr1, n1, arr2, n2));
  
    }
  
}
 
/*This code is contributed by 29AjayKumar*/

PHP

 $max1)
            $max1 = $arr1[$i];
    }
 
    // Find max from 2nd array
    for ($i = 0; $i < $n2; $i++)
    {
        if ($arr2[$i] > $max2)
            $max2 = $arr2[$i];
    }
 
    // Return sum of max from both arrays
    return $max1 + $max2;
}
 
// Driver Code
$arr1 = array( 10, 2, 3 );
 
$arr2 = array( 3, 4, 7 );
 
$n1 = count($arr1);
 
$n2 = count($arr2);
 
echo maxSumPair($arr1, $n1, $arr2, $n2);
 
// This code is contributed by Rajput-Ji
?>

Javascript

 
// Javascript program to find maximum sum
// pair from two arrays
     
    // Function to find maximum sum
    // pair from two arrays
    function maxSumPair(arr1,n1,arr2,n2)
    {
        // max from 1st array
        let max1 = Number.MIN_VALUE;
        // max from 2nd array
        let max2 = Number.MIN_VALUE;
   
     // Find max from 1st array
     for (let i = 0; i < n1; i++) {
         if (arr1[i] > max1)
             max1 = arr1[i];
     }
   
     // Find max from 2nd array
     for (let i = 0; i < n2; i++) {
         if (arr2[i] > max2)
             max2 = arr2[i];
     }
   
     // Return sum of max from both arrays
     return max1 + max2;
    }
     
    // Driver Code
     
    let arr1=[10, 2, 3 ];
    let arr2=[3, 4, 7 ];
    let n1 = arr1.length;
    let n2 = arr2.length;
    document.write(maxSumPair(arr1, n1, arr2, n2));
     
 
// This code is contributed by rag2127
 

输出：

17

时间复杂度：O(N + M)，其中 N 是第一个数组的长度，M 是第二个数组的长度。