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

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

Python3

# Python3 program to find N'th
# element in a set formed
# by sum of two arrays

# Function to calculate the set of sums
def calculateSetOfSum(arr1, size1,arr2, size2, N):

    # Insert each pair sum into set.
    # Note that a set stores elements
    # in sorted order and unique elements
    s = set()
    for i in range(size1):
        for j in range(size2):
            s.add(arr1[i]+arr2[j])

    # If set has less than N elements
    if (len(s) < N):
        return -1

    # Find N'tb item in set and return it
    return list(s)[N - 1]

# Driver code
arr1 = [ 1, 2 ]
size1 = len(arr1)
arr2 = [ 3, 4 ]
size2 = len(arr2)
N = 3

res = calculateSetOfSum(arr1, size1,
                            arr2, size2, N)
if (res == -1):
    print("N'th term doesn't exists in set")
else:
    print(f"N'th element in the set of sums is {res}")


# This code is contributed by shinjanpatra

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

Javascript

<script>

// 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

</script>

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.