在由两个数组之和形成的集合中找到第 N 个项目

给定两个排序数组，我们可以得到一组和（从第一个元素中添加一个元素，从第二个元素中添加一个元素）。查找按排序顺序考虑的已形成集合的元素中的第 N 个元素。
注意：总和集应具有唯一元素。

例子：

Input: arr1[] = {1, 2}
        arr2[] = {3, 4}
        N = 3
Output: 6
We get following elements set of sums.
4(1+3), 5(2+3 or 1+4), 6(2+4)
Third element in above set is 6.

Input: arr1[] = { 1,3, 4, 8, 10} 
        arr2[] = {20, 22, 30, 40} 
        N = 4
Output: 25
We get following elements set of sums.
21(1+20), 23(1+22 or 20+3), 24(20+4), 25(22+3)...
Fourth element is 25.

提问：微软面试

方法：

• 运行两个循环——一个用于第一个数组，第二个用于第二个数组。
• 只需考虑每一对并将它们的总和存储在自平衡 BST 中（由 C++ 中的 set 和 map 实现）。
• 我们在这里在 C++ 中使用 set，因为我们只需要查看元素是否存在，我们不需要键值对。
• 遍历集合并返回集合中的第 N 个元素。

下面是上述方法的实现：

C++

// C++ program to find N'th element in a set formed
// by sum of two arrays
#include
using namespace std;
 
//Function to calculate the set of sums
int calculateSetOfSum(int arr1[], int size1, int arr2[],
                      int size2, int N)
{
    // Insert each pair sum into set. Note that a set
    // stores elements in sorted order and unique elements
    set s;
    for (int i=0 ; i < size1; i++)
        for (int j=0; j < size2; j++)
            s.insert(arr1[i]+arr2[j]);
 
    // If set has less than N elements
    if (s.size() < N)
       return -1;
 
    // Find N'tb item in set and return it
    set::iterator it = s.begin();
    for (int count=1; count

---

Java

// Java program to find N'th element in a set formed
// by sum of two arrays
import java.util.*;
 
class GFG
{
 
// Function to calculate the set of sums
static int calculateSetOfSum(int arr1[], int size1, int arr2[],
                                            int size2, int N)
{
    // Insert each pair sum into set. Note that a set
    // stores elements in sorted order and unique elements
    SortedSet s = new TreeSet();
    for (int i = 0; i < size1; i++)
        for (int j = 0; j < size2; j++)
            s.add(arr1[i]+arr2[j]);
 
    // If set has less than N elements
    if (s.size() < N)
    return -1;
 
    // Find N'tb item in set and return it
     
    return (int)s.toArray()[ N-1 ];
}
 
// Driver code
public static void main(String[] args)
{
    int arr1[] = {1, 2};
    int size1 = arr1.length;
    int arr2[] = {3, 4};
    int size2 = arr2.length;
 
    int N = 3;
    int res = calculateSetOfSum(arr1, size1, arr2, size2, N);
    if (res == -1)
        System.out.println("N'th term doesn't exists in set");
    else
        System.out.println("N'th element in the set of sums is "
            +res);
}
}
 
// This code is contributed by 29AjayKumar

C#

// C# program to find N'th element in
// a set formed by sum of two arrays
using System;
using System.Linq;
using System.Collections.Generic;
     
class GFG
{
 
// Function to calculate the set of sums
static int calculateSetOfSum(int []arr1, int size1,
                             int []arr2, int size2,
                             int N)
{
    // Insert each pair sum into set.
    // Note that a set stores elements in
    // sorted order and unique elements
    HashSet s = new HashSet();
    for (int i = 0; i < size1; i++)
        for (int j = 0; j < size2; j++)
            s.Add(arr1[i] + arr2[j]);
 
    // If set has less than N elements
    if (s.Count < N)
    return -1;
 
    // Find N'tb item in set and return it
    int []last = s.ToArray();
    return last[s.Count - 1];
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr1 = {1, 2};
    int size1 = arr1.Length;
    int []arr2 = {3, 4};
    int size2 = arr2.Length;
 
    int N = 3;
    int res = calculateSetOfSum(arr1, size1,
                                arr2, size2, N);
    if (res == -1)
        Console.WriteLine("N'th term doesn't exists in set");
    else
        Console.WriteLine("N'th element in the set" +
                               " of sums is " + res);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript program to find N'th
// element in a set formed
// by sum of two arrays
 
// Function to calculate the set of sums
function calculateSetOfSum(arr1, size1,
                           arr2, size2, N)
{
     
    // Insert each pair sum into set.
    // Note that a set stores elements
    // in sorted order and unique elements
    let s = new Set();
    for(let i = 0; i < size1; i++)
        for(let j = 0; j < size2; j++)
            s.add(arr1[i]+arr2[j]);
  
    // If set has less than N elements
    if (s.size < N)
        return -1;
  
    // Find N'tb item in set and return it
    return Array.from(s)[N - 1];
}
 
// Driver code
let arr1 = [ 1, 2 ];
let size1 = arr1.length;
let arr2 = [ 3, 4 ];
let size2 = arr2.length;
let N = 3;
 
let res = calculateSetOfSum(arr1, size1,
                            arr2, size2, N);
if (res == -1)
    document.write("N'th term doesn't " +
                   "exists in set");
else
    document.write("N'th element in the set " +
                   "of sums is " + res);
 
// This code is contributed by rag2127
 

输出

N'th element in the set of sums is 6

时间复杂度： O(mn log (mn)) 其中 m 是第一个数组的大小，n 是第二个数组的大小。