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.
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