Circular Matrix (Construct a matrix with numbers 1 to m*n in spiral way)

Given two values m and n, fill a matrix of size ‘m*n’ in a spiral (or circular) fashion (clockwise) with natural numbers from 1 to m*n.

Examples:

Input : m = 4, n = 4
Output :  1  2  3  4
12 13 14  5
11 16 15  6
10  9  8  7

Input : m = 3, n = 4
Output :  1  2  3  4
10 11 12 5
9  8  7  6

The idea is based on Print a given matrix in spiral form. We create a matrix of size m * n and traverse it in a spiral fashion. While traversing, we keep track of a variable “val” to fill the next value, we increment “val” one by one and put its values in the matrix.

Implementation:

C++

 // C++ program to fill a matrix with values from // 1 to n*n in spiral fashion. #include using namespace std;   const int MAX = 100;   // Fills a[m][n] with values from 1 to m*n in // spiral fashion. void spiralFill(int m, int n, int a[][MAX]) {     // Initialize value to be filled in matrix     int val = 1;       /*  k - starting row index         m - ending row index         l - starting column index         n - ending column index */     int k = 0, l = 0;     while (k < m && l < n)     {         /* Print the first row from the remaining           rows */         for (int i = l; i < n; ++i)             a[k][i] = val++;           k++;           /* Print the last column from the remaining           columns */         for (int i = k; i < m; ++i)             a[i][n-1] = val++;         n--;           /* Print the last row from the remaining            rows */         if (k < m)         {             for (int i = n-1; i >= l; --i)                 a[m-1][i] = val++;             m--;         }           /* Print the first column from the remaining            columns */         if (l < n)         {             for (int i = m-1; i >= k; --i)                  a[i][l] = val++;             l++;         }     } }   /* Driver program to test above functions */ int main() {     int m = 4, n = 4;     int a[MAX][MAX];     spiralFill(m, n, a);     for (int i=0; i

C

 // C program to fill a matrix with values from // 1 to n*n in spiral fashion. #include   const int MAX = 100;   // Fills a[m][n] with values from 1 to m*n in // spiral fashion. void spiralFill(int m, int n, int a[][MAX]) {     // Initialize value to be filled in matrix     int val = 1;       /*  k - starting row index         m - ending row index         l - starting column index         n - ending column index */     int k = 0, l = 0;     while (k < m && l < n)     {         /* Print the first row from the remaining           rows */         for (int i = l; i < n; ++i)             a[k][i] = val++;           k++;           /* Print the last column from the remaining           columns */         for (int i = k; i < m; ++i)             a[i][n-1] = val++;         n--;           /* Print the last row from the remaining            rows */         if (k < m)         {             for (int i = n-1; i >= l; --i)                 a[m-1][i] = val++;             m--;         }           /* Print the first column from the remaining            columns */         if (l < n)         {             for (int i = m-1; i >= k; --i)                  a[i][l] = val++;             l++;         }     } }   /* Driver program to test above functions */ int main() {     int m = 4, n = 4;     int a[MAX][MAX];     spiralFill(m, n, a);     for (int i=0; i

Java

 // Java program to fill a matrix with values from // 1 to n*n in spiral fashion. import java.io.*; class GFG {       static int MAX = 100;   // Fills a[m][n] with values from 1 to m*n in // spiral fashion.     static void spiralFill(int m, int n, int a[][]) {         // Initialize value to be filled in matrix         int val = 1;           /*  k - starting row index         m - ending row index         l - starting column index         n - ending column index */         int k = 0, l = 0;         while (k < m && l < n) {             /* Print the first row from the remaining           rows */             for (int i = l; i < n; ++i) {                 a[k][i] = val++;             }               k++;               /* Print the last column from the remaining           columns */             for (int i = k; i < m; ++i) {                 a[i][n - 1] = val++;             }             n--;               /* Print the last row from the remaining            rows */             if (k < m) {                 for (int i = n - 1; i >= l; --i) {                     a[m - 1][i] = val++;                 }                 m--;             }               /* Print the first column from the remaining            columns */             if (l < n) {                 for (int i = m - 1; i >= k; --i) {                     a[i][l] = val++;                 }                 l++;             }         }     }       /* Driver program to test above functions */     public static void main(String[] args) {         int m = 4, n = 4;         int a[][] = new int[MAX][MAX];         spiralFill(m, n, a);         for (int i = 0; i < m; i++) {             for (int j = 0; j < n; j++) {                 System.out.print(a[i][j] + " ");             }             System.out.println("");         }     } }   /* This Java code is contributed by PrinciRaj1992*/

Python3

 # Python program to fill a matrix with # values from 1 to n*n in spiral fashion.   # Fills a[m][n] with values # from 1 to m*n in spiral fashion. def spiralFill(m, n, a):       # Initialize value to be filled in matrix.     val = 1       # k - starting row index     # m - ending row index     # l - starting column index     # n - ending column index     k, l = 0, 0     while (k < m and l < n):           # Print the first row from the remaining rows.         for i in range(l, n):             a[k][i] = val             val += 1         k += 1           # Print the last column from the remaining columns.         for i in range(k, m):             a[i][n - 1] = val             val += 1         n -= 1           # Print the last row from the remaining rows.         if (k < m):             for i in range(n - 1, l - 1, -1):                 a[m - 1][i] = val                 val += 1             m -= 1           # Print the first column from the remaining columns.         if (l < n):             for i in range(m - 1, k - 1, -1):                 a[i][l] = val                 val += 1             l += 1   # Driver program if __name__ == '__main__':     m, n = 4, 4     a = [[0 for j in range(n)] for i in range(m)]     spiralFill(m, n, a)     for i in range(m):         for j in range(n):             print(a[i][j], end=' ')         print('')   # This code is contributed by Parin Shah

C#

 // C# program to fill a matrix with values from // 1 to n*n in spiral fashion.   using System; class GFG {        static int MAX = 100;    // Fills a[m,n] with values from 1 to m*n in // spiral fashion.     static void spiralFill(int m, int n, int[,] a) {         // Initialize value to be filled in matrix         int val = 1;            /*  k - starting row index         m - ending row index         l - starting column index         n - ending column index */         int k = 0, l = 0;         while (k < m && l < n) {             /* Print the first row from the remaining           rows */             for (int i = l; i < n; ++i) {                 a[k,i] = val++;             }                k++;                /* Print the last column from the remaining           columns */             for (int i = k; i < m; ++i) {                 a[i,n - 1] = val++;             }             n--;                /* Print the last row from the remaining            rows */             if (k < m) {                 for (int i = n - 1; i >= l; --i) {                     a[m - 1,i] = val++;                 }                 m--;             }                /* Print the first column from the remaining            columns */             if (l < n) {                 for (int i = m - 1; i >= k; --i) {                     a[i,l] = val++;                 }                 l++;             }         }     }        /* Driver program to test above functions */     public static void Main() {         int m = 4, n = 4;         int[,] a = new int[MAX,MAX];         spiralFill(m, n, a);         for (int i = 0; i < m; i++) {             for (int j = 0; j < n; j++) {                 Console.Write(a[i,j] + " ");             }             Console.Write("\n");         }     } }

PHP

 = \$l; --\$i)                 \$a[\$m - 1][\$i] = \$val++;             \$m--;         }           /* Print the first column from            the remaining columns */         if (\$l < \$n)         {             for (\$i = \$m - 1; \$i >= \$k; --\$i)                 \$a[\$i][\$l] = \$val++;             \$l++;         }     } }   // Driver Code \$m = 4; \$n = 4; spiralFill(\$m, \$n, \$a); for (\$i = 0; \$i < \$m; \$i++) {     for (\$j = 0; \$j < \$n; \$j++)     {         echo (\$a[\$i][\$j]);         echo (" ");     }     echo ("\n"); }   // This code is contributed // by Shivi_Aggarwal ?>

Javascript



Output

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

Time Complexity: O(m * n)
Space Complexity: O(m * n)

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