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Diagonal Number pattern
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Diagonal Number pattern:
Here in this pattern printing, the number of lines of the pattern is taken as the input. Two for loops are used to display the pattern. For each line the number of elements is equal to the number of lines of the pattern. The first for loop (the outer for loop) is used to count the line number so the loop is from 1 to n. For the ith line at the position i the value of i is printed and otherwise 0 is printed.
Algorithm
INPUT: the number of lines
OUTPUT: the aforesaid pattern
PROCESS:
Step 1: read n [the number of lines]
Step 2: for i=1 to n repeat
For j=1 to n repeat
If j=i then
Print i
Else
Print 0
[End of ‘for’ loop]
Move to the next line
[End of ‘for’ loop]
Step 3: stop.
Code
Time Complexity:
for(i=1;i<=n;i++)------------------------------------- n
{ for(j=1;j<=n;j++)--------------------------- n
if(j==i)
printf("%d ",i);
else
printf("0 ");
printf("\n");
}
The complexity is: O(n*n)=O()
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