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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 : 11 Jun, 2021

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 <bits/stdc++.h>
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<Integer> A = new ArrayList<Integer>();
    ArrayList<Integer> B = new ArrayList<Integer>();
     
    // 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

Javascript




<script>
 
    // JavaScript program for the above approach
     
    // Satisfying the given conditions
    function printArrays(n)
    {
 
        // Declare the two arrays A and B
        let A = [];
        let B = [];
 
        // Iterate from range [1, 2*n]
        for(let i = 1; i <= 2 * n; i++)
        {
 
            // Assign consecutive numbers to
            // same indices of the two arrays
            if (i % 2 == 0)
                A.push(i);
            else
                B.push(i);
        }
 
        // Print the first array
        document.write("{ ");
        for(let i = 0; i < n; i++)
        {
            document.write(A[i]);
 
            if (i != n - 1)
                document.write(", ");
        }
        document.write(" }" + "</br>");
 
        // Print the second array, B
        document.write("{ ");
        for(let i = 0; i < n; i++)
        {
            document.write(B[i]);
 
            if (i != n - 1)
                document.write(", ");
        }
        document.write(" }");
    }
     
    let N = 5;
      
    // Function Call
    printArrays(N);
 
</script>
Output: 
{ 2, 4, 6, 8, 10 }
{ 1, 3, 5, 7, 9 }

 

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

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