Find N'th item in a set formed by sum of two arrays
Last Updated :
13 Sep, 2023
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 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 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# 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
<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>
OutputN'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.
Similar Reads
Maximum sum by picking elements from two arrays in order | Set 2 Given two arrays A[] and B[], each of size N, and two integers X and Y denoting the maximum number of elements that can be picked from A[] and B[] respectively, the task is to find the maximum possible sum by selecting N elements in such a way that for any index i, either A[i] or B[i] can be chosen.
10 min read
Find sub-arrays from given two arrays such that they have equal sum Given two arrays A[] and B[] of equal sizes i.e. N containing integers from 1 to N. The task is to find sub-arrays from the given arrays such that they have equal sum. Print the indices of such sub-arrays. If no such sub-arrays are possible then print -1.Examples: Input: A[] = {1, 2, 3, 4, 5}, B[] =
15+ min read
Find the overlapping sum of two arrays Given two arrays A[] and B[] having n unique elements each. The task is to find the overlapping sum of the two arrays. That is the sum of elements that is common in both of the arrays. Note: Elements in the arrays are unique. That is the array does not contain duplicates. Examples: Input : A[] = {1,
8 min read
Minimum sum of two elements from two arrays such that indexes are not same Given two arrays a[] and b[] of same size. Task is to find minimum sum of two elements such that they belong to different arrays and are not at same index in their arrays. Examples: Input : a[] = {5, 4, 13, 2, 1} b[] = {2, 3, 4, 6, 5}Output : 3We take 1 from a[] and 2 from b[]Sum is 1 + 2 = 3.Input
15+ min read
Javascript Program for Find k pairs with smallest sums in two arrays Given two integer arrays arr1[] and arr2[] sorted in ascending order and an integer k. Find k pairs with smallest sums such that one element of a pair belongs to arr1[] and other element belongs to arr2[]Examples: Input : arr1[] = {1, 7, 11} arr2[] = {2, 4, 6} k = 3Output : [1, 2], [1, 4], [1, 6]Exp
3 min read