Form an array of distinct elements with each element as sum of an element from each array

Given two arrays, arr1[] and arr2[] of equal size, which contain distinct elements, the task is to find another array with distinct elements such that elements of the third array are formed by the addition of the one-one element of the arr1[] and arr2[].

Examples:  

Input: arr[] = {1, 7, 8, 3}, arr2[] = {6, 5, 10, 2} 
Output: 3 8 13 18 
Explanation: 
Index 0: 1 + 2 = 3 
Index 1: 3 + 5 = 8 
Index 2: 7 + 6 = 13 
Index 3: 8 + 10 = 18 
The elements of the array are distinct. 

Input: arr1[] = {15, 20, 3}, arr2[] = {5, 4, 3} 
Output: 6 19 25 
Explanation: 
Index 0: 3 + 3 = 6 
Index 1: 15 + 4 = 19 
Index 2: 20 + 5 = 25 
The elements of the array are distinct. 

Approach: The key observation in this problem is that both arrays contain distinct elements and if we sort the array, then the sum of corresponding elements of the array will also form distinct elements.
The step-by-step algorithm for the above approach is described below- 

• Sort both the arrays in an increasing or decreasing order.
• Initialize another array(say arr3[]) to store the distinct elements formed by the sum of two elements of the array
• Iterate over a loop from 0 to the length of the array
• Elements of the third array will be the sum of the elements of the first two arrays-

 arr3[i] = arr1[i] + arr2[i]

Below is the implementation of the above approach: 

// C++ implementation to find distinct
// array such that the elements of the
// array is sum of two 
// elements of other two arrays

#include <bits/stdc++.h>
using namespace std;

// Function to find a distinct array
// such that elements of the third 
// array is sum of two elements 
// of other two arrays 
void thirdArray(int a[], int b[], int n)
{
    int c[n];
    sort(a, a + n);
    sort(b, b + n);
    
    // Loop to find the array 
    // such that each element is
    // sum of other two elements
    for (int i = 0; i < n; i++) {
        c[i] = a[i] + b[i];
    }
    
    // Loop to print the array
    for (int i = 0; i < n; i++)
        cout << c[i] << " ";
}

// Driver code
int main()
{
    int a[] = { 1, 7, 8, 3 };
    int b[] = { 6, 5, 10, 2 };

    int size = sizeof(a) / sizeof(a[0]);

    thirdArray(a, b, size);

    return 0;
}

// Java implementation to find distinct
// array such that the elements of the
// array is sum of two 
// elements of other two arrays
import java.util.*; 

class GFG
{
    // Function to find a distinct array
    // such that elements of the third 
    // array is sum of two elements 
    // of other two arrays 
    static void thirdArray(int a[], int b[], int n)
    {
        int[] c = new int[20];;
        Arrays.sort(a);
        Arrays.sort(b);
        
        // Loop to find the array 
        // such that each element is
        // sum of other two elements
        for (int i = 0; i < n; i++) {
            c[i] = a[i] + b[i];
        }
        
        // Loop to print the array
        for (int i = 0; i < n; i++)
            System.out.print(c[i] + " ");
    }
    
    // Driver code
    public static void main(String args[]) 
    {
        int a[] = { 1, 7, 8, 3 };
        int b[] = { 6, 5, 10, 2 };
    
        int size = a.length;
    
        thirdArray(a, b, size);
    }
}

// This code is contributed by shubhamsingh10

# Python3 implementation to find distinct
# array such that the elements of the
# array is sum of two
# elements of other two arrays

# Function to find a distinct array
# such that elements of the third
# array is sum of two elements
# of other two arrays
def thirdArray(a, b, n):

    c = [0]*n
    a = sorted(a)
    b = sorted(b)

    # Loop to find the array
    # such that each element is
    # sum of other two elements
    for i in range(n):
        c[i] = a[i] + b[i]

    # Loop to print the array
    for i in range(n):
        print(c[i], end=" ")

# Driver code
a = [1, 7, 8, 3]
b = [6, 5, 10, 2]

size = len(a)

thirdArray(a, b, size)

# This code is contributed by mohit kumar 29    

// C# implementation to find distinct
// array such that the elements of the
// array is sum of two 
// elements of other two arrays
using System;

class GFG
{
    // Function to find a distinct array
    // such that elements of the third 
    // array is sum of two elements 
    // of other two arrays 
    static void thirdArray(int []a, int []b, int n)
    {
        int[] c = new int[20];;
        Array.Sort(a);
        Array.Sort(b);
         
        // Loop to find the array 
        // such that each element is
        // sum of other two elements
        for (int i = 0; i < n; i++) {
            c[i] = a[i] + b[i];
        }
         
        // Loop to print the array
        for (int i = 0; i < n; i++)
            Console.Write(c[i] + " ");
    }
     
    // Driver code
    public static void Main(String []args) 
    {
        int []a = { 1, 7, 8, 3 };
        int []b = { 6, 5, 10, 2 };
     
        int size = a.Length;
     
        thirdArray(a, b, size);
    }
}

// This code is contributed by PrinciRaj1992

<script>

// javascript implementation to find distinct
// array such that the elements of the
// array is sum of two 
// elements of other two arrays

// Function to find a distinct array
// such that elements of the third 
// array is sum of two elements 
// of other two arrays 
function thirdArray(a, b, n)
{
    var c = new Array(n);
    a = a.sort(function(a, b) {
          return a - b;
    });
    b = b.sort(function(a, b) {
          return a - b;
    });
    
    var i,j;
    // Loop to find the array 
    // such that each element is
    // sum of other two elements
    for (i = 0; i < n; i++) {
        c[i] = a[i] + b[i];
    }
    
    // Loop to print the array
    for (i = 0; i < n; i++)
        document.write(c[i] + " ");
}

// Driver code

    var a = [1, 7, 8, 3];
    var b = [6, 5, 10, 2];

    var size = a.length;
    thirdArray(a, b, size);

</script>

3 8 13 18

Performance Analysis:  

• Time Complexity: As in the above approach, there is sorting the array of size N which takes O(N*logN) time, Hence the Time Complexity will be O(N*logN).
• Auxiliary Space: As in the above approach, extra space for array c is being used, Hence the space complexity will be O(N).