Related Articles
Construct two N-length arrays with same-indexed elements as co-prime and a difference of N in their sum
• Difficulty Level : Expert
• Last Updated : 10 Dec, 2020

Given a positive integer N, the task is to generate two arrays of length N such that the same-indexed elements of both the arrays are co-prime and absolute difference between the sum of elements of the arrays is N.

Examples:

Input: N = 5
Output:
{1, 3, 5, 7, 9}
{2, 4, 6, 8, 10}
Explanation: Pairs of same-indexed elements are (1, 2), (3, 4), (5, 6), (7, 8), (9, 10). It can be observed that all the pairs are coprime and the absolute difference of the sum of the two arrays is 5.

Input: N = 3
Output:
{2, 4, 7}
{1, 3, 6}

Approach: The idea is based on the observation that two consecutive natural numbers are always co-prime and the difference between them is 1. Follow the steps below to solve the problem:

• Initialize two arrays A[] and B[] of size N.
• Iterate over the range [1, 2*N] using the variable i. For every element in the range, check if i is divisible by 2 or not. If found to be true, then insert i into the array A[]. Otherwise, insert i into the array B[].
• After completing the above steps, print the two arrays.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to generate two arrays``// satisfying the given conditions``void` `printArrays(``int` `n)``{``    ``// Declare the two arrays A and B``    ``vector<``int``> A, B;` `    ``// Iterate from range [1, 2*n]``    ``for` `(``int` `i = 1; i <= 2 * n; i++) {` `        ``// Assign consecutive numbers to``        ``// same indices of the two arrays``        ``if` `(i % 2 == 0)``            ``A.push_back(i);``        ``else``            ``B.push_back(i);``    ``}` `    ``// Print the first array``    ``cout << ``"{ "``;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``cout << A[i];` `        ``if` `(i != n - 1)``            ``cout << ``", "``;``    ``}``    ``cout << ``" }\n"``;` `    ``// Print the second array, B``    ``cout << ``"{ "``;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``cout << B[i];` `        ``if` `(i != n - 1)``            ``cout << ``", "``;``    ``}``    ``cout << ``" }"``;``}` `// Driver Code``int` `main()``{``    ``int` `N = 5;` `    ``// Function Call``    ``printArrays(N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.util.*;`` ` `class` `GFG{``      ` `// Satisfying the given conditions``static` `void` `printArrays(``int` `n)``{``    ` `    ``// Declare the two arrays A and B``    ``ArrayList A = ``new` `ArrayList();``    ``ArrayList B = ``new` `ArrayList();``    ` `    ``// Iterate from range [1, 2*n]``    ``for``(``int` `i = ``1``; i <= ``2` `* n; i++)``    ``{``        ` `        ``// Assign consecutive numbers to``        ``// same indices of the two arrays``        ``if` `(i % ``2` `== ``0``)``            ``A.add(i);``        ``else``            ``B.add(i);``    ``}`` ` `    ``// Print the first array``    ``System.out.print(``"{ "``);``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``        ``System.out.print(A.get(i));`` ` `        ``if` `(i != n - ``1``)``            ``System.out.print(``", "``);``    ``}``    ``System.out.print(``" }\n"``);`` ` `    ``// Print the second array, B``    ``System.out.print(``"{ "``);``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``        ``System.out.print(B.get(i));``        ` `        ``if` `(i != n - ``1``)``            ``System.out.print(``", "``);``    ``}``    ``System.out.print(``" }"``);``}`` ` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `N = ``5``;``    ` `    ``// Function Call``    ``printArrays(N);``}``}` `// This code is contributed by sanjoy_62`

## Python3

 `# Python3 program for the above approach` `# Function to generate two arrays``# satisfying the given conditions``def` `printArrays(n) :``    ` `    ``# Declare the two arrays A and B``    ``A, B ``=` `[], [];` `    ``# Iterate from range [1, 2*n]``    ``for` `i ``in` `range``(``1``, ``2` `*` `n ``+` `1``):``        ` `        ``# Assign consecutive numbers to``        ``# same indices of the two arrays``        ``if` `(i ``%` `2` `=``=` `0``) :``            ``A.append(i);``        ``else` `:``            ``B.append(i);` `    ``# Print the first array``    ``print``(``"{ "``, end``=``"");``    ``for` `i ``in` `range``(n) :``        ``print``(A[i], end``=``"");` `        ``if` `(i !``=` `n ``-` `1``) :``            ``print``(``", "``, end``=``"");``    ` `    ``print``(``"}"``);` `    ``# Print the second array, B``    ``print``(``"{ "``, end``=``"");``    ` `    ``for` `i ``in` `range``(n) :``        ``print``(B[i], end``=``"");` `        ``if` `(i !``=` `n ``-` `1``) :``            ``print``(``","``, end``=``" "``);``            ` `    ``print``(``" }"``, end``=``"");` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``N ``=` `5``;` `    ``# Function Call``    ``printArrays(N);` `    ``# This code is contributed by AnkitRai01`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{``  ` `// Satisfying the given conditions``static` `void` `printArrays(``int` `n)``{``    ` `    ``// Declare the two arrays A and B``    ``List<``int``> A = ``new` `List<``int``>();``    ``List<``int``> B = ``new` `List<``int``>();``     ` `    ``// Iterate from range [1, 2*n]``    ``for``(``int` `i = 1; i <= 2 * n; i++)``    ``{``        ` `        ``// Assign consecutive numbers to``        ``// same indices of the two arrays``        ``if` `(i % 2 == 0)``            ``A.Add(i);``        ``else``            ``B.Add(i);``    ``}``  ` `    ``// Print the first array``    ``Console.Write(``"{ "``);``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ``Console.Write(A[i]);``  ` `        ``if` `(i != n - 1)``            ``Console.Write(``", "``);``    ``}``    ``Console.Write(``" }\n"``);``  ` `    ``// Print the second array, B``    ``Console.Write(``"{ "``);``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ``Console.Write(B[i]);``         ` `        ``if` `(i != n - 1)``            ``Console.Write(``", "``);``    ``}``    ``Console.Write(``" }"``);``}``  ` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `N = 5;``    ` `    ``// Function Call``    ``printArrays(N);``}``}` `// This code is contributed by susmitakundugoaldanga`
Output:
```{ 2, 4, 6, 8, 10 }
{ 1, 3, 5, 7, 9 }```

Time Complexity: O(N)
Auxiliary Space: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up