此实现计数排序的时间复杂度

Question

I have to determine the time complexity for the given code of counting sort. I know that counting sort should be O(n+k) but I just can't see how this function is O(n+k), seeing as there is a for loop inside the second for loop.

def counting_sort(array, k): # k is max(array)

    count = (k+1) * [0]  
    sorted = []
    for i in array:
        count[i] += 1
    for i in range(len(count)):
        if count[i] != 0:
            sorted += [i for j in range(count[i])]
    return sorted

Wouldn't the worst case be n+k^2, if the elements are unique?

Answer 1

The first loop

for i in array:
    count[i] += 1

takes n iterations, for the second loop the number of instructions executed in the worst case scenario for the list comprehension:

i for j in range(count[i]) #

is count[i] , and the sum of all count[i] is equal to n. Note that the comparison

        if count[i] != 0:

is done k times, and in the worst case scenario the

sorted +=...

is also done k-times. Adding all this up you get your O(n+k)

(assuming the += for sorted is constant cost, otherwise we have to say that the += is amortized constant and so the result comes with that caveat).