Question

I have read a counting sort algorithm which is like this:

Counting Sort(A[1,..n]) //C[1,...k] is the temporary memory and k is the range of integers
   for  i<-- 1 to k
      C[i]<-- 0
   for  j<-- 1 to n
      C[A[j]]<--C[A[j]]+1
   for  i<--2 to k
      C[i]<--C[i]+C[i-1]
   for  j<--n downto 1
      B[C[A[j]]]<--A[j]
      C[A[j]]<--C[A[j]]-1

I want to know that if I change the last for to this:for j<--1 to n ,the algorithm will be correct too???(is there any way to show that with this “for” the algorithm will be correct??? )

also in this way the algorithm is stable too?

thanks

Answer 1

The algorithm is correct both ways. It is also stable as you have it right now.

If you change the last for to what you said, it will stop being stable.

Basically, C[i] = how many elements <= i exist after the end of the third for loop. So C[A[j]] gives you the the last position of an element with value A[j] in sorted order, C[A[j]] - 1 the second last position of an element with value A[j] and so on. This is why you decrement the values in C.

Because of this, you have to start iterating your original array in reverse order if you care about stability: so that the last element with value x in your original array gets put first in the new array. Iterating your original array in reverse will guarantee that x is put after all other values equal to x, thus making the algorithm stable.