Given a R x C matrix Fill matrix with values from 1 to RxC spirally. We’ve to repeatedly print elements using these orders (Left->Right, Top->Down, Right->Left, Down->Up)
Input Format
Line 1: Integer T - number of test cases.
Next T lines contain a pair of integers per line R C
Constraints
1 <= T <= 100
1 <= R, C <= 100Output Format
For each test case, print RxC matrix on output which is filled spirally. Also print newline after all the elements have been printed.
Sample Input 0
2
3 3
2 4Sample Output 0
1 2 3
8 9 4
7 6 5
1 2 3 4
8 7 6 5Solution in java
import java.util.Scanner;
class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int num_of_cases = sc.nextInt();
while (num_of_cases-- > 0) {
int r = sc.nextInt();
int c = sc.nextInt();
int num = 1;
int up = 0;
int left = 0;
int down = r - 1;
int right = c - 1;
int matrix[][] = new int[r][c];
while (num <= r * c) {
for (int col = left; col <= right; col++) {
matrix[up][col] = num;
num++;
}
for (int row = up + 1; row <= down; row++) {
matrix[row][right] = num;
num++;
}
if (up != down) {
for (int col = right - 1; col >= left; col--) {
matrix[down][col] = num;
num++;
}
}
if (left != right) {
for (int row = down - 1; row > up; row--) {
matrix[row][left] = num;
num++;
}
}
left++;
right--;
up++;
down--;
}
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
System.out.print(matrix[i][j] + " ");
}
System.out.println();
}
System.out.println();
}
sc.close();
}
}