## Solid Half Diamond Left

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### Description

Solid Half Diamond left:

Here the number of lines is taken as input. The aforesaid pattern is divided into two halves while implemented. In the first half, the numbers of stars are increased, the spaces before the stars are decreased and in the second half, the number of stars decreased, the numbers of spaces are increased. For each half two loops are used to print the stars. The numbers of stars are increased by 1 and in the second half, the numbers of stars are decreased by 1. The pattern is only possible if the number of lines is odd.

### Algorithm

``````INPUT: the number of lines
OUTPUT: the aforesaid pattern
PROCESS:
Step 1: [taking the inputs]
Step 2: [printing the pattern]
Set st<-1
Set sp<-n/2
If n modulus 2≠0 then
For i=1 to n/2+1 repeat
For k=1 to sp repeat
Print “ “
[End of ‘for’ loop]
For j=1 to st repeat
Print "*"
[End of ‘for’ loop]
Move to the next line
Set st<-st+1
Set sp<-sp-1
[End of ‘for’ loop]
Set st<-st-2
Set sp<-sp+2
For i=n/2+2 to n repeat
For k=1 to sp repeat
Print “ “
[End of ‘for’ loop]
For j=1 to st repeat
Print "*"
[End of ‘for’ loop]
Move to the next line
Set st<-st-1
Set sp<-sp+1
[End of ‘for’ loop]
else
print "Please give correct input(no. of lines should be odd)
[End of ‘if’]
Step 3: Stop
``````

## Time Complexity:

for(i=1;i<=n/2+1;i++)-----------------------------------------------------O(n/2+1)

{     for(k=1;k<=sp;k++)---------------------------------------------O(sp)

printf(" ");

for(j=1;j<=st;j++)------------------------------------------------O(st)

printf("*");

printf("\n");

sp--;

st+=1;   }

st=st-2;

sp+=2;

for(i=n/2+2;i<=n;i++)-----------------------------------------------O(n/2)

{              for(k=1;k<=sp;k++)----------------------------------------O(sp)

printf(" ");

for(j=1;j<=st;j++)------------------------------------------O(st)

printf("*");

printf("\n");

st-=1;

sp+=1;  }

The complexity of this program=O((n/2+1)*(sp+st)+ (n/2)*(sp+st))=O($\sqrt{\mathrm{n}}$).