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C Diamond Star Pattern
Why Star Pattern
In most of the MNC interview questions such as in ZOHO interview questions, IVTL Infoview interview questions, Amazon interview questions, GOOGLE interview questions, Infosys interview questions and even in Voonik interview questions, we come across star pattern questions with a probability of 1:5. We provide you a most commonly asked Diamond star pattern question for your interview preparation using C programming.
Lets code a C program to print diamond star pattern to have some fun
Diamond Star Pattern
We challenge you, the programmer to do the diamond star pattern using c programming
C Programming logic
1) Split this program into 2 parts.
2) part one contains.
* *** ***** ******* *********
3) part two contains.
*******
*****
***
*
Diamond Star Pattern Program - Part 1
Consider every single space in part 1 as 0.
0000* 000*** 00***** 0******* *********
#include<stdio.h>
int main()
{
int i, j, k;
for(i = 1;i <= 5;i++)
{
for(j = 5;j > i;j--)
{
printf("0");
}
for(k = 1;k <= ((j*2)-1);k++)
{
printf("*");
}
printf("\n");
}
return 0;
}
0000* 000*** 00***** 0******* *********
Diamond Star Pattern - Part 2
0******* 00***** 000*** 0000*
#include<stdio.h>
int main()
{
int i, j, k, m = 8;
for(i = 1;i <= 4;i++)
{
for(j = 1;j <= i;j++)
{
printf("0");
}
for(k = i;k < m;k++)
{
printf("*");
}
m-=1;
printf("\n");
}
return 0;
}
0******* 00***** 000*** 0000*
Diamond Star Pattern - Combined
0000* 000*** 00***** 0******* ********* 0******* 00***** 000*** 0000*
#include<stdio.h>
int main()
{
int i, j, k, m = 8;
for(i = 1;i <= 5;i++)
{
for(j = 5;j > i;j--)
{
printf("0");
}
for(k = 1;k <= ((j*2)-1);k++)
{
printf("*");
}
printf("\n");
}
for(i = 1;i <= 4;i++)
{
for(j = 1;j <= i;j++)
{
printf("0");
}
for(k = i;k < m;k++)
{
printf("*");
}
m-=1;
printf("\n");
}
return 0;
}
0000* 000*** 00***** 0******* ********* 0******* 00***** 000*** 0000*
Diamond Star Pattern - Complete
*
***
*****
*******
*********
*******
*****
***
*
#include<stdio.h>
int main()
{
int i, j, k, m = 8;
for(i = 1;i <= 5;i++)
{
for(j = 5;j > i;j--)
{
printf(" ");
}
for(k = 1;k <= ((j*2)-1);k++)
{
printf("*");
}
printf("\n");
}
for(i = 1;i <= 4;i++)
{
for(j = 1;j <= i;j++)
{
printf(" ");
}
for(k = i;k < m;k++)
{
printf("*");
}
m-=1;
printf("\n");
}
return 0;
}
*
***
*****
*******
*********
*******
*****
***
*
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