Form coils in a matrix

Given a positive integer n which represents the dimensions of a 4n x 4n matrix with values from 1 to n filled from left to right and top to bottom. Form two coils from the matrix and print the coils.

Examples: 

Input  : n = 1;
Output : Coil 1 : 10 6 2 3 4 8 12 16 
         Coil 2 : 7 11 15 14 13 9 5 1
Explanation : Matrix is 
1  2  3  4 
5  6  7  8 
9  10 11 12 
13 14 15 16

Input  : n = 2;
Output : Coil 1 : 36 28 20 21 22 30 38 46 54 
                  53 52 51 50 42 34 26 18 10 
                  2 3 4 5 6 7 8 16 24 32 40 
                  48 56 64 
        Coil 2 : 29 37 45 44 43 35 27 19 11 12 
                 13 14 15 23 31 39 47 55 63 62 
                 61 60 59 58 57 49 41 33 25 17
                 9 1 

The total elements in the matrix are 16n². All elements are divided into two coils. Every coil has 8n² elements. We make two arrays of this size. We first fill elements in coil1 by traversing them in the given order. Once we have filled elements in coil1, we can get elements of other coil2 using formula coil2[i] = 16*n*n + 1 -coil1[i]. 

Implementation:

C++

// C++ program to print 2 coils of a
// 4n x 4n matrix.
#include<iostream>
using namespace std;

// Print coils in a matrix of size 4n x 4n
void printCoils(int n)
{
    // Number of elements in each coil
    int m = 8*n*n;

    // Let us fill elements in coil 1.
    int coil1[m];

    // First element of coil1
    // 4*n*2*n + 2*n;
    coil1[0] = 8*n*n + 2*n;
    int curr = coil1[0];

    int nflg = 1, step = 2;

    // Fill remaining m-1 elements in coil1[]
    int index = 1;
    while (index < m)
    {
        // Fill elements of current step from
        // down to up
        for (int i=0; i<step; i++)
        {
            // Next element from current element
            curr = coil1[index++] = (curr - 4*n*nflg);
            if (index >= m)
                break;
        }
        if (index >= m)
            break;

        // Fill elements of current step from
        // up to down.
        for (int i=0; i<step; i++)
        {
            curr = coil1[index++] = curr + nflg;
            if (index >= m)
                break;
        }
        nflg = nflg*(-1);
        step += 2;
    }

    /* get coil2 from coil1 */
    int coil2[m];
    for (int i=0; i<8*n*n; i++)
        coil2[i] = 16*n*n + 1 -coil1[i];

    // Print both coils
    cout << "Coil 1 : ";
    for(int i=0; i<8*n*n; i++)
        cout << coil1[i] << " ";
    cout << "\nCoil 2 : ";
    for (int i=0; i<8*n*n; i++)
        cout << coil2[i] << " ";
}

// Driver code
int main()
{
    int n = 1;
    printCoils(n);
    return 0;
}

Java

// Java program to print 2 coils
// of a 4n x 4n matrix.

class GFG {

    // Print coils in a matrix of size 4n x 4n
    static void printCoils(int n)
    {
        // Number of elements in each coil
        int m = 8 * n * n;

        // Let us fill elements in coil 1.
        int coil1[] = new int[m];

        // First element of coil1
        // 4*n*2*n + 2*n;
        coil1[0] = 8 * n * n + 2 * n;
        int curr = coil1[0];

        int nflg = 1, step = 2;

        // Fill remaining m-1 elements in coil1[]
        int index = 1;
        while (index < m)
        {
            // Fill elements of current step from
            // down to up
            for (int i = 0; i < step; i++)
                {
                    // Next element from current element
                    curr = coil1[index++] = (curr - 4 * n * nflg);
                    if (index >= m)
                    break;
                }
            if (index >= m)
                break;

            // Fill elements of current step from
            // up to down.
            for (int i = 0; i < step; i++)
            {
                curr = coil1[index++] = curr + nflg;
                if (index >= m)
                break;
            }

            nflg = nflg * (-1);
            step += 2;
        }

        /* get coil2 from coil1 */
        int coil2[] = new int[m];
        for (int i = 0; i < 8 * n * n; i++)
            coil2[i] = 16 * n * n + 1 - coil1[i];

        // Print both coils
        System.out.print("Coil 1 : ");

        for (int i = 0; i < 8 * n * n; i++)
            System.out.print(coil1[i] + " ");

        System.out.print("\nCoil 2 : ");

        for (int i = 0; i < 8 * n * n; i++)
            System.out.print(coil2[i] + " ");
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 1;
        printCoils(n);
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python3 program to print 2 coils of a
# 4n x 4n matrix.

# Print coils in a matrix of size 4n x 4n
def printCoils(n):

    # Number of elements in each coil
    m = 8*n*n

    # Let us fill elements in coil 1.
    coil1 = [0]*m

    # First element of coil1
    # 4*n*2*n + 2*n
    coil1[0] = 8*n*n + 2*n

    curr = coil1[0]

    nflg = 1
    step = 2

    # Fill remaining m-1 elements in coil1[]
    index = 1
    while (index < m):

        # Fill elements of current step from
        # down to up
        for i in range(step):

            # Next element from current element
            curr = coil1[index] = (curr - 4*n*nflg)
            index += 1
            if (index >= m):
                break
        if (index >= m):
            break

        # Fill elements of current step from
        # up to down.
        for i in range(step):

            curr = coil1[index] = curr + nflg
            index += 1
            if (index >= m):
                break
        nflg = nflg*(-1)
        step += 2

    #get coil2 from coil1 */

    coil2 = [0]*m
    i = 0
    while(i < 8*n*n):
        coil2[i] = 16*n*n + 1 -coil1[i]
        i += 1
    # Print both coils
    print("Coil 1 :", end = " ")
    i = 0
    while(i < 8*n*n):
        print(coil1[i], end = " ")
        i += 1
    print("\nCoil 2 :", end = " ")
    i = 0
    while(i < 8*n*n):
        print(coil2[i], end = " ")
        i += 1

# Driver code

n = 1
printCoils(n)

# This code is contributed by shubhamsingh10

C#

// C# program to print 2 coils
// of a 4n x 4n matrix.
using System;

class GFG {

    // Print coils in a matrix of
    // size 4n x 4n
    static void printCoils(int n)
    {

        // Number of elements in
        // each coil
        int m = 8 * n * n;

        // Let us fill elements in
        // coil 1.
        int [] coil1 = new int[m];

        // First element of coil1
        // 4*n*2*n + 2*n;
        coil1[0] = 8 * n * n + 2 * n;
        int curr = coil1[0];

        int nflg = 1, step = 2;

        // Fill remaining m-1 elements
        // in coil1[]
        int index = 1;
        while (index < m)
        {

            // Fill elements of current
            // step from down to up
            for (int i = 0; i < step; i++)
                {
                    // Next element from
                    // current element
                    curr = coil1[index++]
                    = (curr - 4 * n * nflg);

                    if (index >= m)
                        break;
                }

            if (index >= m)
                break;

            // Fill elements of current step
            // from up to down.
            for (int i = 0; i < step; i++)
            {
                curr = coil1[index++] = curr
                                     + nflg;
                if (index >= m)
                    break;
            }

            nflg = nflg * (-1);
            step += 2;
        }

        /* get coil2 from coil1 */
        int [] coil2 = new int[m];

        for (int i = 0; i < 8 * n * n; i++)
            coil2[i] = 16 * n * n + 1 - coil1[i];

        // Print both coils
        Console.Write("Coil 1 : ");

        for (int i = 0; i < 8 * n * n; i++)
            Console.Write(coil1[i] + " ");

        Console.Write("\nCoil 2 : ");

        for (int i = 0; i < 8 * n * n; i++)
            Console.Write(coil2[i] + " ");
    }

    // Driver code
    public static void Main()
    {
        int n = 1;

        printCoils(n);
    }
}

// This code is contributed by KRV.

PHP

<?php
// PHP program to print 2 coils of a
// 4n x 4n matrix.

// Print coils in a matrix of size 4n x 4n
function printCoils( $n)
{

    // Number of elements in each coil
    $m = 8 * $n * $n;

    // Let us fill elements in coil 1.
    $coil1 = array();

    // First element of coil1
    // 4*n*2*n + 2*n;
    $coil1[0] = 8 * $n * $n + 2 * $n;
    $curr = $coil1[0];

    $nflg = 1; $step = 2;

    // Fill remaining m-1 elements in coil1[]
    $index = 1;
    while ($index < $m)
    {
        // Fill elements of current step from
        // down to up
        for ( $i = 0; $i < $step; $i++)
        {
            // Next element from current element
            $curr = $coil1[$index++] =
                     ($curr - 4*$n*$nflg);
            if ($index >= $m)
                break;
        }
        if ($index >= $m)
            break;

        // Fill elements of current step from
        // up to down.
        for ( $i=0; $i<$step; $i++)
        {
            $curr = $coil1[$index++] =
                            $curr + $nflg;
            if ($index >= $m)
                break;
        }
        $nflg = $nflg * (-1);
        $step += 2;
    }

    /* get coil2 from coil1 */
    $coil2 = array();

    for ( $i = 0; $i < 8 * $n * $n; $i++)
        $coil2[$i] = 16 * $n * $n + 1 -$coil1[$i];

    // Print both coils
    echo "Coil 1 : ";
    for( $i = 0; $i < 8 * $n * $n; $i++)
    echo $coil1[$i] , " ";

    echo "\nCoil 2 : ";
    for ( $i = 0; $i < 8 * $n * $n; $i++)
        echo $coil2[$i] , " ";
}

// Driver code
$n = 1;
printCoils($n);

// This code is contributed by anuj_67.
?>

Javascript

<script>
    // Javascript program to print 2 coils
    // of a 4n x 4n matrix.

    // Print coils in a matrix of
    // size 4n x 4n
    function printCoils(n)
    {

        // Number of elements in
        // each coil
        let m = 8 * n * n;

        // Let us fill elements in
        // coil 1.
        let coil1 = new Array(m);
        coil1.fill(0);

        // First element of coil1
        // 4*n*2*n + 2*n;
        coil1[0] = 8 * n * n + 2 * n;
        let curr = coil1[0];

        let nflg = 1, step = 2;

        // Fill remaining m-1 elements
        // in coil1[]
        let index = 1;
        while (index < m)
        {

            // Fill elements of current
            // step from down to up
            for (let i = 0; i < step; i++)
                {
                    // Next element from
                    // current element
                    curr = coil1[index++]
                    = (curr - 4 * n * nflg);

                    if (index >= m)
                        break;
                }

            if (index >= m)
                break;

            // Fill elements of current step
            // from up to down.
            for (let i = 0; i < step; i++)
            {
                curr = coil1[index++] = curr
                                     + nflg;
                if (index >= m)
                    break;
            }

            nflg = nflg * (-1);
            step += 2;
        }

        /* get coil2 from coil1 */
        let coil2 = new Array(m);
        coil2.fill(0);

        for (let i = 0; i < 8 * n * n; i++)
            coil2[i] = 16 * n * n + 1 - coil1[i];

        // Print both coils
        document.write("Coil 1 : ");

        for (let i = 0; i < 8 * n * n; i++)
            document.write(coil1[i] + " ");

        document.write("</br>" + "Coil 2 : ");

        for (let i = 0; i < 8 * n * n; i++)
            document.write(coil2[i] + " ");
    }

    let n = 1;

      printCoils(n);

</script>

Output

Coil 1 : 10 6 2 3 4 8 12 16 
Coil 2 : 7 11 15 14 13 9 5 1 

Time Complexity: O(n²)
Auxiliary Space: O(n²)

Asked in Yahoo