Latest C Aptitude Question SOLUTION: 4 4 4 4 4 4 4 4 3 3 3 3 3 4 4 3 2 2 2 3 4 4 3 2 1 2 3 4 4

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(#M40165556) C Programming question print the pattern of the following form containing the numbers from 1to n Keep an EYE

4 4 4 4 4 4 4
4 3 3 3 3 3 4
4 3 2 2 2 3 4
4 3 2 1 2 3 4
4 3 2 2 2 3 4
4 3 3 3 3 3 4
4 4 4 4 4 4 4

Asked In C Sannapureddy Srinivasareddy (5 years ago)

Unsolved Read Solution (5)

Is this Puzzle helpful?   (21)   (1) Submit Your Solution Program

Read Solution (5) :

ritesh  (5 years ago)

#include
int main()
{
int n,i,j,k;
printf("enter number nn");
scanf("%d",&n);
for(i=1;i

Is this solution Helpfull? Yes (7) | No (21)  

ritesh  (5 years ago)

#include
void main()
{
int i,j,k,n;
printf("enter n");
scanf("%d",&n);
for(i=1;i

Is this solution Helpfull? Yes | No (9)  

  (5 years ago)

#include
int main()
{
int n=4,a;
a=n*2-1;
for(int i=0;i

Is this solution Helpfull? Yes | No (2)  

Varinderjit Singh  (4 years ago)

Get the solution at:
https://drive.google.com/file/d/16FkcQr9TM1IHx52uhHEOYdp6iIxzwIve/view?usp=drivesdk

Is this solution Helpfull? Yes | No   

  (4 years ago)

// Function to print the pattern
void printPattern(int n)
{
int arraySize = n * 2 - 1;
int result[arraySize][arraySize];

// Fill the values
for(int i = 0; i < arraySize; i++)
{
for(int j = 0; j < arraySize; j++)
{
if(abs(i - arraySize / 2) > abs(j - arraySize / 2))
result[i][j] = abs(i - arraySize / 2) + 1;
else
result[i][j] = (abs(j-arraySize / 2) + 1);

}
}

// Print the array
for(int i = 0; i < arraySize; i++)
{
for(int j = 0; j < arraySize; j++)
{
cout

Is this solution Helpfull? Yes | No   

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