Find N'th item in a set formed by sum of two arrays

Find N’th item in a set formed by sum of two arrays

Difficulty Level : Medium
Last Updated : 25 Sep, 2020

Given two sorted arrays, we can get a set of sums(add one element from the first and one from second). Find the N’th element in the elements of the formed set considered in sorted order.
Note: Set of sums should have unique elements.
Examples:

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.

Asked in: Microsoft Interview

Approach:

Run two loops – one for the first array and second for the second array.
Just consider each pair and store their sum in a self-balancing-BST (which is implemented by set and map in C++).
We use set in C++ here as we need to only see if elements are present or absent, we don’t need key, value pairs.
Traverse the set and return the Nth element in the set.

Below is the implementation of the above approach:

C++

// C++ program to find N'th element in a set formed
// by sum of two arrays
#include<bits/stdc++.h>
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<int> 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<int>::iterator it = s.begin();
    for (int count=1; count<N; count++)
        it++;
    return *it;
}
 
// Driver code
int main()
{
    int arr1[] = {1, 2};
    int size1 = sizeof(arr1) / sizeof(arr1[0]);
    int arr2[] = {3, 4};
    int size2 = sizeof(arr2) / sizeof(arr2[0]);
 
    int N = 3;
    int res = calculateSetOfSum(arr1, size1, arr2, size2, N);
    if (res == -1)
        cout << "N'th term doesn't exists in set";
    else
        cout << "N'th element in the set of sums is "
             << res;
    return 0;
}

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<Integer> s = new TreeSet<Integer>();
    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<int> s = new HashSet<int>();
    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

Output

N'th element in the set of sums is 6

Time Complexity: O(mn log (mn)) where m is the size of the first array and n is the size of the second array.

This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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