Half Right Diamond Number Pattern:
Here for printing the pattern the number of lines is taken as inputs. The pattern is divided into two halves. In the first half the number of elements is increased and in the 2nd half the number of elements is decreased. For ith line the difference between two consecutive elements is i. the number of lines of this pattern should be odd.
INPUT: the number of lines
OUTPUT: the aforesaid pattern
PROCESS:
Step 1: read n [number of lines]
Step 2: [Printing the pattern]
If n mod 2≠0 then
For i=1 to n/2+1 repeat
Set k<-i
For j=1 to i repeat
Print k
Set k<-k+i
Move to the next line
[End of ‘for’ loop]
Set e<-i-2
For i=n/2+2 to n repeat
Set k<-i
For j=1 to e repeat
Print k
Set k<-k+i
[End of ‘for’ loop]
Move to the next line
Set e<-e-1
[End of ‘for’ loop]
else
print "The pattern can be printed only if the number of lines is odd"
[End of ‘if’]
Step 3. Stop.
for(i=1;i<=n/2+1;i++)----------------------------------------------------------------- n/2+1
{ for(j=1,k=i;j<=i;j++,k+=i)----------------------- i
{ printf("%d ",k);
}
printf("\n");
}
e=i-2;
for(i=n/2+2;i<=n;i++)---------------------------------------- n/2
{ for(j=1,k=i;j<=e;j++,k+=i)------------------------ e
{ printf("%d ",k);
}
printf("\n");
e--;
}
The complexity is: O(((n/2+1)*i)+(n/2*e))=O()
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