Objective
In this problem, you need to print the pattern of the following form containing the numbers from 1 to n.
4 4 4 4 4 4 44 3 3 3 3 3 44 3 2 2 2 3 44 3 2 1 2 3 44 3 2 2 2 3 44 3 3 3 3 3 44 4 4 4 4 4 4
Input Format
The input will contain a single integer n.
Constraints
1<=n<=1000
Output Format
Print the pattern mentioned in the problem statement.
Example 0 :
Sample Input 0
2
Sample Output 0
2 2 2 2 1 2 2 2 2
Example 1 :
Sample Input 1
5
Sample Output 1
5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 5 5 4 3 3 3 3 3 4 5 5 4 3 2 2 2 3 4 5 5 4 3 2 1 2 3 4 5 5 4 3 2 2 2 3 4 5 5 4 3 3 3 3 3 4 5 5 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5
Example 2 :
Sample Input 2
7
Sample Output 2
7 7 7 7 7 7 7 7 7 7 7 7 7 7 6 6 6 6 6 6 6 6 6 6 6 7 7 6 5 5 5 5 5 5 5 5 5 6 7 7 6 5 4 4 4 4 4 4 4 5 6 7 7 6 5 4 3 3 3 3 3 4 5 6 7 7 6 5 4 3 2 2 2 3 4 5 6 7 7 6 5 4 3 2 1 2 3 4 5 6 7 7 6 5 4 3 2 2 2 3 4 5 6 7 7 6 5 4 3 3 3 3 3 4 5 6 7 7 6 5 4 4 4 4 4 4 4 5 6 7 7 6 5 5 5 5 5 5 5 5 5 6 7 7 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7
Printing Pattern Using Loops – Hacker Rank Solution
Code:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main()
{
int i,j,k,m,n,x;
scanf("%d",&n);
k=n;
m = n+(n-1);
for(i=0;i<m;i++)
{
for(j=0;j<m;j++)
{
if(i<=n-1)
{
if(i==0)
{
printf("%d ",k);
}
if(i>=1)
{
if(j<i)
{
printf("%d ",k-j);
}
else if(j>=i && j<m-i)
{
printf("%d ",k-i);
}
else
{
printf("%d ",(j-k+1)+1);
}
}
}
else if(i==n-1)
{
if(j<n)
{
printf("%d ",k-j);
}
else
{
printf("%d ",(j-k+1)+1);
}
}
else if(i>=n)
{
x = m-i-1;
if(i==m)
{
printf("%d ",k);
}
if(j<x)
{
printf("%d ",k-j);
}
else if(j>=x && j<m-x)
{
printf("%d ",k-x);
}
else
{
printf("%d ",(j-k+1)+1);
}
}
}
printf("\n");
}
}
Disclaimer: The above Problem (Printing Pattern Using Loops ) is generated by Hackerrank but the Solution is Provided by MultiplexCoder. This tutorial is only for Educational and Learning purposes.
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