C Mirror Half 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 mirror half diamond star pattern question for your interview preparation using C programming.

Mirror Half Diamond Star Pattern

We challenge you, the programmer to do the mirror half diamond star pattern using C programming

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    *

C Programming logic

Lets code a C program to print mirror half diamond star pattern to have some fun

step 1: Split this program into 2 parts.

step 2: Part one contains.

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*****

step 3: Part two contains.

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   *

step 4: From the above star pattern, we can easily identify this c program contains 3 for loop.

• for loop represents every new line.
• for loop for displaying 0.
• for loop for displaying *.

step 5: Finally, combine these two star pattern.

Facts

• C program is not capable of transferring its control from line 2 to line 1.
• Thus the programmer need to complete the line 1 before executing the line 2.

Mirror Half Diamond Star Pattern - Part 1

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#include<stdio.h>
int main()
{
int i, j, k;
for(i = 1;i <= 5;i++)
{
for(j = 5;j > i;j--)
{
printf(" ");
}
for(k = 1;k <= j;k++)
{
printf("*");
}
printf("\n");
}
return 0;
}

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   **
  ***
 ****
*****

Mirror Half Diamond Star Pattern - Part 2

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 ***
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   *

#include<stdio.h>
int main()
{
int i, j, k;
for(i = 1;i <= 4;i++)
{
for(j = 1;j < i;j++)
{
printf(" ");
}
for(k = 4;k >= i;k--)
{
printf("*");
}
printf("\n");
}
return 0;
}

****
 ***
  **
   *

Mirror Half Diamond Star Pattern

    *
   **
  ***
 ****
*****
 ****
  ***
   **
    *

#include<stdio.h>
int main()
{
int i, j, k;
for(i = 1;i <= 5;i++)
{
for(j = 5;j > i;j--)
{
printf(" ");
}
for(k = 1;k <= j;k++)
{
printf("*");
}
printf("\n");
}
for(i = 1;i <= 4;i++)
{
for(j = 1;j <= i;j++)
{
printf(" ");
}
for(k = 4;k >= i;k--)
{
printf("*");
}
printf("\n");
}
return 0;
}

    *
   **
  ***
 ****
*****
 ****
  ***
   **
    *

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