从给定的数组中选择两个元素，以使它们的和不出现在任何数组中

给定两个数组A []和B [] ，任务是选择两个元素X和Y ，使得X属于A []且Y属于B []，并且(X + Y)不能存在于任何一个大批。
例子：

Input: A[] = {3, 2, 2}, B[] = {1, 5, 7, 7, 9} 
Output: 3 9 
3 + 9 = 12 and 12 is not present in 
any of the given arrays.
Input: A[] = {1, 3, 5, 7}, B[] = {7, 5, 3, 1} 
Output: 7 7 

方法：选择X作为从A []和Y从B []的最大元素作为最大元素。现在，很明显(X + Y)将大于两个数组的最大值，即它不会出现在所有数组中。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to find the numbers from
// the given arrays such that their
// sum is not present in any
// of the given array
void findNum(int a[], int n, int b[], int m)
{
    // Find the maximum element
    // from both the arrays
    int x = *max_element(a, a + n);
    int y = *max_element(b, b + m);
    cout << x << " " << y;
}
 
// Driver code
int main()
{
    int a[] = { 3, 2, 2 };
    int n = sizeof(a) / sizeof(int);
    int b[] = { 1, 5, 7, 7, 9 };
    int m = sizeof(b) / sizeof(int);
 
    findNum(a, n, b, m);
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
// find maximum element in an array
static int max_element(int a[], int n)
{
    int m = Integer.MIN_VALUE;
     
    for(int i = 0; i < n; i++)
        m = Math.max(m, a[i]);
     
    return m;
}
 
// Function to find the numbers from
// the given arrays such that their
// sum is not present in any
// of the given array
static void findNum(int a[], int n,
                    int b[], int m)
{
    // Find the maximum element
    // from both the arrays
    int x = max_element(a, n);
    int y = max_element(b, m);
    System.out.print(x + " " + y);
}
 
// Driver code
public static void main(String args[])
{
    int a[] = { 3, 2, 2 };
    int n = a.length;
    int b[] = { 1, 5, 7, 7, 9 };
    int m = b.length;
 
    findNum(a, n, b, m);
}
}
 
// This code is contributed by Arnub Kundu

Python3

# Python3 implementation of the approach
 
# Function to find the numbers from
# the given arrays such that their
# sum is not present in any
# of the given array
def findNum(a, n, b, m) :
 
    # Find the maximum element
    # from both the arrays
    x = max(a);
    y = max(b);
    print(x, y);
 
# Driver code
if __name__ == "__main__" :
 
    a = [ 3, 2, 2 ];
    n = len(a);
     
    b = [ 1, 5, 7, 7, 9 ];
    m = len(b);
 
    findNum(a, n, b, m);
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
    // find maximum element in an array
    static int max_element(int []a, int n)
    {
        int m = int.MinValue;
         
        for(int i = 0; i < n; i++)
            m = Math.Max(m, a[i]);
         
        return m;
    }
     
    // Function to find the numbers from
    // the given arrays such that their
    // sum is not present in any
    // of the given array
    static void findNum(int []a, int n,
                        int []b, int m)
    {
        // Find the maximum element
        // from both the arrays
        int x = max_element(a, n);
        int y = max_element(b, m);
        Console.Write(x + " " + y);
    }
     
    // Driver code
    public static void Main()
    {
        int []a = { 3, 2, 2 };
        int n = a.Length;
        int []b = { 1, 5, 7, 7, 9 };
        int m = b.Length;
     
        findNum(a, n, b, m);
    }
}
 
// This code is contributed by kanugargng

Javascript

 
// Javascript implementation of the approach
 
// Function to find the numbers from
// the given arrays such that their
// sum is not present in any
// of the given array
function findNum(a, n, b, m)
{
    // Find the maximum element
    // from both the arrays
    var x = a.reduce(function(a, b) { return Math.max(a, b); });
    var y = b.reduce(function(a, b) { return Math.max(a, b); });
    document.write(x + " " + y);
}
 
// Driver code
var a = [ 3, 2, 2 ];
var n = a.length;
var b = [ 1, 5, 7, 7, 9 ]
var m = b.length;
findNum(a, n, b, m);
 
// This code is contributed by rutvik_56.

输出：

3 9

时间复杂度： O(n)