# Spirally traversing a matrix

QUESTION

Traverse a 4×4 matrix of integers in spiral form.\n\nInput: The first line of input contains an integer T denoting the number of test cases. First four lines of the test case will contain four elements each.\nOutput: Spiral array will be displayed in a single line.\nConstraints:\n\n1 <=T<= 100\n\n1 <=N<= 100.

“TESTCASE_1”: “1\n1 2 3 4\n5 6 7 8\n9 10 11 12\n13 14 15 16\n###—###SEPERATOR—###—\n1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10”, “TESTCASE_2”: “1\n1 5 9 13\n2 6 10 14\n3 7 11 15\n4 8 12 16\n###—###SEPERATOR—###—\n1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7”, “TESTCASE_3”: “0\n###—###SEPERATOR—###—\n0”, “TESTCASE_4”: “0\n###—###SEPERATOR—###—\n0”, “TESTCASE_5”: “0\n###—###SEPERATOR—###—\n0

``````#include <stdio.h>
#define R 4
#define C 4

void spiralPrint(int m, int n, int a[R][C])
{
int i, k = 0, l = 0;

while (k < m && l < n)
{

for (i = l; i < n; ++i)
{
printf("%d ", a[k][i]);
}
k++;

for (i = k; i < m; ++i)
{
printf("%d ", a[i][n-1]);
}
n--;

if ( k < m)
{
for (i = n-1; i >= l; --i)
{
printf("%d ", a[m-1][i]);
}
m--;
}

if (l < n)
{
for (i = m-1; i >= k; --i)
{
printf("%d ", a[i][l]);
}
l++;
}
}
}

int main()
{
int t;
scanf("%d",&t);
while(t!=0)
{
int a[4][4],i,j;
for(i=0;i<4;i++)
{
for(j=0;j<4;j++)
{
scanf("%d",&a[i][j]);
}
}

spiralPrint(R, C, a);
t--;
}
return 0;
}``````