Printing Pattern using Loops - Hacker rank Solution
Problem
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 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 Input Format
Constraints
1<=n<=1000Output Format
Print the pattern mentioned in the problem statement.
Sample Output 2
Disclaimer :-
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
Solution :-
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 | #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 whole problem statement are given by Hackerrank.com but the solution are generated by the codeworld19 authority if any of the query regarding this post or website fill the following contact form thank you.
CodeWorld19
Hey, I am Parthe Omkar
This solution is also work
#include
#include
#include
#include
int main()
{
int n;
scanf("%d", &n);
for(int i = n; i >= 1; i--){
for(int j = n; j > i; j--)
printf("%d ", j);
for(int j = 1; j <= 2 * i - 1; j++)
printf("%d ", i);
for(int j = i + 1; j <= n; j++)
printf("%d ", j);
printf("\n");
}
for(int i = 1; i < n; i++){
for(int j = n; j > i; j--)
printf("%d ", j);
for(int j = 1; j <= 2 * i - 1; j++)
printf("%d ", i + 1);
for(int j = i + 1; j <= n; j++)
printf("%d ", j);
printf("\n");
}
return 0;
}
Not working
Works thank you
Hello everyone I have not sure about the solution which has provide by some one in comment section. follow the solution which I have given in post section
-if any one has occurred problem then comment, otherwise fill the following contact form.
Thank you 🙏
This comment has been removed by the author.
can you also explain the logic, i have a hard time solving this kind of looping patterns in c. or at least can you give me tips to solve any kind of looping patterns in c? ugh i have a bad logic
#include
#include
#include
#include
int main()
{
int i,j,k,m,n,x;
scanf("%d",&n); //3
k=n; //2
m = n+(n-1); //m=5
for(i=0;i=1) //i=2 j=4
{
if(j=i && j=i j<5-i
{
printf("%d ",k-i);
}
else
{
printf("%d ",(j-k+1)+1);
}
}
}
else if(i>=n) //i>=3 i=4 j=2 [k=3,m=5,n=3,x=4-i]
{
x = m-i-1; //x=0 j=2
if(j=x && j<m-x)
{
printf("%d ",k-x);
}
else
{
printf("%d ",(j-k+1)+1);
}
}
}
}
return 0;
}
#include
#include
using namespace std;
int main()
{
int n;
cin>>n;
for (int i = -n + 1; i < n; i++)
{
for (int j = -n + 1; j < n; j++)
{
cout<<" "<<max(abs(i), abs(j)) + 1;
}
cout<<"\n";
}
return 0;
}
how to print it
#include
int main()
{
int n;
scanf("%d",&n);
int size=2*n-1;
int start=0;
int end=size-1;
int a [size][size];
while(n != 0)
{
for(int i=start;i<=end;i++)
{
for(int j=start;j<=end;j++)
{
if(i==start || i==end || j==start || j ==end)
a[i][j];
}
}
++ start;
--end;
--n;
}
return 0;
}
galat h code
This comment has been removed by the author.
we design promotional USBs
#include
#include
#include
#include
int main()
{
int n,i,j,line=0,N,element=0;
scanf("%d", &n);N=2*n-1;
for(i=0;in-line)
element--;
}
else
{
if(j+line>N-2)
element++;
}
}printf("\n");
if(i<N/2)
{
line++;
}
else
{
line--;
}
}
return 0;
}
This comment has been removed by the author.
This comment has been removed by the author.
public class bigMatrix {
public static void main(String args[])
{
int n=4;
for(int row=0;row<=n*2;row++)
{
int colNum=(row>n)?(2*n-row):row;
for(int col=0;col<=2*n;col++)
{
if(col<=colNum )
System.out.print(col);
else if ((n*2-col)<=colNum)
System.out.print(n*2-col+" hi ");
else
System.out.print(colNum);
}
System.out.println();
}
}
}