Given two arrays arr1[] and arr2[] of equal size which contains 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
Explantion:
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 is 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 is 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 such in increasing or decreasing order.
- Intialize another array(say arr3[]) to store the distinct elements formed by sum of two elements of the array
- Iterate over a loop from 0 to 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 prthe 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 |
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).
- Space Complexity: As in the above approach, there is no extra space used, Hence the space complexity will be O(1).
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