There are balls of red, blue and green color. How will you sort them…?
I just wanted to know the best stratergy for it…!
Thanks in advance.

use counting sort…

just put some value to each color
 i.e. red=0; blue=1;green=2;
example ; red green blue look like 0 2 1

and just take a hash[3]={0}
for(all balls not taken){ if(ball==red) hash[0]++; else if(ball==blue) hash[1]++; else hash[2]++; }

you will got to know how many red ball and how many blue and how many green

in which order you want put them as you know the frequency of each colored ball…

just do this way…to more info of counting sort…Counting sort
HAPPY CODING
it means after counting the frequency and then arranging them in some order (say red first, then blue, then green), minimum 2n comparisons required at least .!
Hey va1ts7_100,
firstly you should give different value to red, green and blue balls.
Suppose there are n number of balls.
We define array of of size 3 and initialize them to 0.
red=0
blue=1
green=2
while(n–)
{
scanf(ball color)
array[ball color]++
}
for(i=0 to 2)
{
for(j=1 to array[ball color])
{
if(i==0)
print(red ball)
else `if(i==1)
print(blue ball)
else
print(green ball)
}
}
i think above code should help you!
thanks@deepkmourya
just give them values 0 ,1 and 2
then sort them using any nlogn complexity sort like merge sort…
or you can use the inbuilt sort function if you are using c++…
in this type of question firstly you should check if other way to sort which has less complexity than n*log(n) , if it clicks in your mind your code take less time as compared to others.
HAAPY CODING

**ok if you want to efficient than do one thing make an empty array ***

assumption ie. red=0,blue=1,green=2;

make three variable
 r=0;
 g=totalnumberofballs1;
 b=g1;
put[totalnumber of balls]//empty list while(all ball not scanned){ if(ball==red){ put[r]=red; r++; } else if(ball==green){ if(put[g]==blue){ put[b]=blue; b } put[g]=green; g; } else { put[b]=blue; b; } }
HAPPY CODING
ONLY n COMPARISION IS NEEDED…HOPE IT HELP YOU…
i could have done the same using 2 arrays by myself, but i was trying to achieve it in only 1 array…!! thanks for ur solution…
 you can do this also …
 i will post that solution soon…just wait…
 i came to know that solution too…
just wait…
 or you have to do some shift but it take more time than this…
 i will see…