Java Program to Form coils in a matrix

13 Jun, 2022
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]. 
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("Coil 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.
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), where n represents the given integer.Auxiliary Space: O(n2), where n represents the given integer.
Please refer complete article on Form coils in a matrix for more details!
tamanna17122007
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