Number Pattern:
Here the number of lines is taken as input. The aforesaid pattern is divided into two halves while implemented. The first half the numbers of elements are increased and in the second half the number of elements decreased. For each half two for loops are used to print the elements. The numbers of elements are increased by 1 and in the second half the numbers of elements are decreased by 1. The pattern is only possible if the number of lines is odd.
INPUT: Number of lines
OUTPUT: the aforesaid pattern
PROCESS:
Step 1: [taking the inputs]
Read n [number of lines]
Step 2: [printing the pattern]
Set st<-1
Set d<-n/2
If n modulus 2≠0 then
For i=1 to n/2+1 repeat
For j=1 to st repeat
Print d
[End of ‘for’ loop]
Move to the next line
Set st<-st+1
Set d<-d+1
[End of ‘for’ loop]
Set st<-st-2
Set d<-d-2
For i=n/2+2 to n repeat
For j=1 to st repeat
Print d
[End of ‘for’ loop]
Move to the next line
Set st<-st-1
Set d<-d-1
[End of ‘for’ loop]
else
print "Please give correct input(no. of lines should be odd)
[End of ‘if’]
Step 3: Stop
for(i=1;i<=n/2+1;i++)--------------------------------------------------------------- n/2+1
{ for(j=1;j<=st;j++)------------------------------- st
printf("%d",d);
printf("\n");
st+=1;
d++;
}
st=st-2;
d-=2;
for(i=n/2+2;i<=n;i++)-------------------------------------------- n/2
{ for(j=1;j<=st;j++)---------------------------------------- st
printf("%d",d);
printf("\n");
st-=1;
d--;
}
}
The complexity is: O(((n/2+1)*st)+((n/2)*st))=O()
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