Number Pattern :
The number of lines is taken as input. For each line, the values are incremented and the values are the square of consecutive integers. Three for loops are used to print the pattern. The outer for loop is used to count the number of lines. The two inner loops are used to print the spaces and elements of each line.
INPUT: The 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
Set q<-1
For i=1 to n repeat
For k=1 to n-i repeat
Print “ “
[End of ‘for’ loop]
For j=1 to i repeat
Print p2
Set p<-p+1
[End of ‘for’ loop]
Set q<-q+2
[End of ‘for’ loop]
Step 3: Stop.
for(i=1;i<=n;i++)------------------------------------------------------ n
{ for(k=1;k<=n-i;k++)-------------------------- n-i
printf(" ");
for(j=1;j<=q;j++)------------------------------ q
{ printf("%d ",(int)pow(p,2));
p=p+1; }
q=q+2;
printf("\n"); }
The complexity is: O(n*(n-i+q))=O()
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