Number Pattern 13:
For printing the aforesaid pattern, the number of lines is taken as input. For the odd lines the values are printed in descending order and for the even lines the values are printed in ascending order.
INPUT: Number of lines
OUTPUT: the aforesaid pattern
PROCESS:
Step 1: [taking the input]
Read n [number of lines]
Step 2: [printing the pattern]
Set p<-1
For i=1 to n repeat
If i mod 2=0 then
For j=1 to i repeat
Print p
Set p<-p+1
[End of ‘for’ loop]
else
Set p<-(p+i)-1
For j=1 to i repeat
Print p
Set p<-p-1
[End of ‘for’ loop]
Set p<-p+i+1
[End of if]
Move to the next line
[End of ‘for’ loop]
Step 3: Stop.
for(i=1;i<=n;i++)-----------------------------------------------------------n
{ if(i%2==0)
{ for(j=1;j<=i;j++)---------------------------------------i
{ printf("%d ",p);
p++; }
}
else
{ p=(p+i)-1;
for(j=1;j<=i;j++)----------------------------------------i
{ printf("%d ",p);
p--; }
p=p+i+1; }
printf("\n");
}
The complexity is: O(n*(i+i))=O()
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