'Quicksort with Python
I am totally new to python and I am trying to implement quicksort in it. Could someone please help me complete my code?
I do not know how to concatenate the three arrays and printing them.
def sort(array=[12,4,5,6,7,3,1,15]):
less = []
equal = []
greater = []
if len(array) > 1:
pivot = array[0]
for x in array:
if x < pivot:
less.append(x)
if x == pivot:
equal.append(x)
if x > pivot:
greater.append(x)
sort(less)
sort(pivot)
sort(greater)
Solution 1:[1]
def sort(array=[12,4,5,6,7,3,1,15]):
"""Sort the array by using quicksort."""
less = []
equal = []
greater = []
if len(array) > 1:
pivot = array[0]
for x in array:
if x < pivot:
less.append(x)
elif x == pivot:
equal.append(x)
elif x > pivot:
greater.append(x)
# Don't forget to return something!
return sort(less)+equal+sort(greater) # Just use the + operator to join lists
# Note that you want equal ^^^^^ not pivot
else: # You need to handle the part at the end of the recursion - when you only have one element in your array, just return the array.
return array
Solution 2:[2]
Quick sort without additional memory (in place)
Usage:
array = [97, 200, 100, 101, 211, 107]
quicksort(array)
print(array)
# array -> [97, 100, 101, 107, 200, 211]
def partition(array, begin, end):
pivot = begin
for i in range(begin+1, end+1):
if array[i] <= array[begin]:
pivot += 1
array[i], array[pivot] = array[pivot], array[i]
array[pivot], array[begin] = array[begin], array[pivot]
return pivot
def quicksort(array, begin=0, end=None):
if end is None:
end = len(array) - 1
def _quicksort(array, begin, end):
if begin >= end:
return
pivot = partition(array, begin, end)
_quicksort(array, begin, pivot-1)
_quicksort(array, pivot+1, end)
return _quicksort(array, begin, end)
Solution 3:[3]
There is another concise and beautiful version
def qsort(arr):
if len(arr) <= 1:
return arr
else:
return qsort([x for x in arr[1:] if x < arr[0]])
+ [arr[0]]
+ qsort([x for x in arr[1:] if x >= arr[0]])
Let me explain the above codes for details
pick the first element of array
arr[0]as pivot[arr[0]]qsortthose elements of array which are less than pivot withList Comprehensionqsort([x for x in arr[1:] if x < arr[0]])qsortthose elements of array which are larger than pivot withList Comprehensionqsort([x for x in arr[1:] if x >= arr[0]])
Solution 4:[4]
This answer is an in-place QuickSort for Python 2.x. My answer is an interpretation of the in-place solution from Rosetta Code which works for Python 3 too:
import random
def qsort(xs, fst, lst):
'''
Sort the range xs[fst, lst] in-place with vanilla QuickSort
:param xs: the list of numbers to sort
:param fst: the first index from xs to begin sorting from,
must be in the range [0, len(xs))
:param lst: the last index from xs to stop sorting at
must be in the range [fst, len(xs))
:return: nothing, the side effect is that xs[fst, lst] is sorted
'''
if fst >= lst:
return
i, j = fst, lst
pivot = xs[random.randint(fst, lst)]
while i <= j:
while xs[i] < pivot:
i += 1
while xs[j] > pivot:
j -= 1
if i <= j:
xs[i], xs[j] = xs[j], xs[i]
i, j = i + 1, j - 1
qsort(xs, fst, j)
qsort(xs, i, lst)
And if you are willing to forgo the in-place property, below is yet another version which better illustrates the basic ideas behind quicksort. Apart from readability, its other advantage is that it is stable (equal elements appear in the sorted list in the same order that they used to have in the unsorted list). This stability property does not hold with the less memory-hungry in-place implementation presented above.
def qsort(xs):
if not xs: return xs # empty sequence case
pivot = xs[random.choice(range(0, len(xs)))]
head = qsort([x for x in xs if x < pivot])
tail = qsort([x for x in xs if x > pivot])
return head + [x for x in xs if x == pivot] + tail
Solution 5:[5]
Quicksort with Python
In real life, we should always use the builtin sort provided by Python. However, understanding the quicksort algorithm is instructive.
My goal here is to break down the subject such that it is easily understood and replicable by the reader without having to return to reference materials.
The quicksort algorithm is essentially the following:
- Select a pivot data point.
- Move all data points less than (below) the pivot to a position below the pivot - move those greater than or equal to (above) the pivot to a position above it.
- Apply the algorithm to the areas above and below the pivot
If the data are randomly distributed, selecting the first data point as the pivot is equivalent to a random selection.
Readable example:
First, let's look at a readable example that uses comments and variable names to point to intermediate values:
def quicksort(xs):
"""Given indexable and slicable iterable, return a sorted list"""
if xs: # if given list (or tuple) with one ordered item or more:
pivot = xs[0]
# below will be less than:
below = [i for i in xs[1:] if i < pivot]
# above will be greater than or equal to:
above = [i for i in xs[1:] if i >= pivot]
return quicksort(below) + [pivot] + quicksort(above)
else:
return xs # empty list
To restate the algorithm and code demonstrated here - we move values above the pivot to the right, and values below the pivot to the left, and then pass those partitions to same function to be further sorted.
Golfed:
This can be golfed to 88 characters:
q=lambda x:x and q([i for i in x[1:]if i<=x[0]])+[x[0]]+q([i for i in x[1:]if i>x[0]])
To see how we get there, first take our readable example, remove comments and docstrings, and find the pivot in-place:
def quicksort(xs):
if xs:
below = [i for i in xs[1:] if i < xs[0]]
above = [i for i in xs[1:] if i >= xs[0]]
return quicksort(below) + [xs[0]] + quicksort(above)
else:
return xs
Now find below and above, in-place:
def quicksort(xs):
if xs:
return (quicksort([i for i in xs[1:] if i < xs[0]] )
+ [xs[0]]
+ quicksort([i for i in xs[1:] if i >= xs[0]]))
else:
return xs
Now, knowing that and returns the prior element if false, else if it is true, it evaluates and returns the following element, we have:
def quicksort(xs):
return xs and (quicksort([i for i in xs[1:] if i < xs[0]] )
+ [xs[0]]
+ quicksort([i for i in xs[1:] if i >= xs[0]]))
Since lambdas return a single epression, and we have simplified to a single expression (even though it is getting more unreadable) we can now use a lambda:
quicksort = lambda xs: (quicksort([i for i in xs[1:] if i < xs[0]] )
+ [xs[0]]
+ quicksort([i for i in xs[1:] if i >= xs[0]]))
And to reduce to our example, shorten the function and variable names to one letter, and eliminate the whitespace that isn't required.
q=lambda x:x and q([i for i in x[1:]if i<=x[0]])+[x[0]]+q([i for i in x[1:]if i>x[0]])
Note that this lambda, like most code golfing, is rather bad style.
In-place Quicksort, using the Hoare Partitioning scheme
The prior implementation creates a lot of unnecessary extra lists. If we can do this in-place, we'll avoid wasting space.
The below implementation uses the Hoare partitioning scheme, which you can read more about on wikipedia (but we have apparently removed up to 4 redundant calculations per partition() call by using while-loop semantics instead of do-while and moving the narrowing steps to the end of the outer while loop.).
def quicksort(a_list):
"""Hoare partition scheme, see https://en.wikipedia.org/wiki/Quicksort"""
def _quicksort(a_list, low, high):
# must run partition on sections with 2 elements or more
if low < high:
p = partition(a_list, low, high)
_quicksort(a_list, low, p)
_quicksort(a_list, p+1, high)
def partition(a_list, low, high):
pivot = a_list[low]
while True:
while a_list[low] < pivot:
low += 1
while a_list[high] > pivot:
high -= 1
if low >= high:
return high
a_list[low], a_list[high] = a_list[high], a_list[low]
low += 1
high -= 1
_quicksort(a_list, 0, len(a_list)-1)
return a_list
Not sure if I tested it thoroughly enough:
def main():
assert quicksort([1]) == [1]
assert quicksort([1,2]) == [1,2]
assert quicksort([1,2,3]) == [1,2,3]
assert quicksort([1,2,3,4]) == [1,2,3,4]
assert quicksort([2,1,3,4]) == [1,2,3,4]
assert quicksort([1,3,2,4]) == [1,2,3,4]
assert quicksort([1,2,4,3]) == [1,2,3,4]
assert quicksort([2,1,1,1]) == [1,1,1,2]
assert quicksort([1,2,1,1]) == [1,1,1,2]
assert quicksort([1,1,2,1]) == [1,1,1,2]
assert quicksort([1,1,1,2]) == [1,1,1,2]
Conclusion
This algorithm is frequently taught in computer science courses and asked for on job interviews. It helps us think about recursion and divide-and-conquer.
Quicksort is not very practical in Python since our builtin timsort algorithm is quite efficient, and we have recursion limits. We would expect to sort lists in-place with list.sort or create new sorted lists with sorted - both of which take a key and reverse argument.
Solution 6:[6]
There are many answers to this already, but I think this approach is the most clean implementation:
def quicksort(arr):
""" Quicksort a list
:type arr: list
:param arr: List to sort
:returns: list -- Sorted list
"""
if not arr:
return []
pivots = [x for x in arr if x == arr[0]]
lesser = quicksort([x for x in arr if x < arr[0]])
greater = quicksort([x for x in arr if x > arr[0]])
return lesser + pivots + greater
You can of course skip storing everything in variables and return them straight away like this:
def quicksort(arr):
""" Quicksort a list
:type arr: list
:param arr: List to sort
:returns: list -- Sorted list
"""
if not arr:
return []
return quicksort([x for x in arr if x < arr[0]]) \
+ [x for x in arr if x == arr[0]] \
+ quicksort([x for x in arr if x > arr[0]])
Solution 7:[7]
functional approach:
def qsort(lst):
if len(lst) < 2:
return lst
pivot = lst[0]
left = list(filter(lambda x: x <= pivot, lst[1:]))
right = list(filter(lambda x: x > pivot, lst[1:]))
return qsort(left) + [pivot] + qsort(right)
Solution 8:[8]
Easy implementation from grokking algorithms
def quicksort(arr):
if len(arr) < 2:
return arr #base case
else:
pivot = arr[0]
less = [i for i in arr[1:] if i <= pivot]
more = [i for i in arr[1:] if i > pivot]
return quicksort(less) + [pivot] + quicksort(more)
Solution 9:[9]
functional programming aproach
smaller = lambda xs, y: filter(lambda x: x <= y, xs)
larger = lambda xs, y: filter(lambda x: x > y, xs)
qsort = lambda xs: qsort(smaller(xs[1:],xs[0])) + [xs[0]] + qsort(larger(xs[1:],xs[0])) if xs != [] else []
print qsort([3,1,4,2,5]) == [1,2,3,4,5]
Solution 10:[10]
This is a version of the quicksort using Hoare partition scheme and with fewer swaps and local variables
def quicksort(array):
qsort(array, 0, len(array)-1)
def qsort(A, lo, hi):
if lo < hi:
p = partition(A, lo, hi)
qsort(A, lo, p)
qsort(A, p + 1, hi)
def partition(A, lo, hi):
pivot = A[lo]
i, j = lo-1, hi+1
while True:
i += 1
j -= 1
while(A[i] < pivot): i+= 1
while(A[j] > pivot ): j-= 1
if i >= j:
return j
A[i], A[j] = A[j], A[i]
test = [21, 4, 1, 3, 9, 20, 25, 6, 21, 14]
print quicksort(test)
Solution 11:[11]
Partition - Split an array by a pivot that smaller elements move to the left and greater elemets move to the right or vice versa. A pivot can be an random element from an array. To make this algorith we need to know what is begin and end index of an array and where is a pivot. Then set two auxiliary pointers L, R.
So we have an array user[...,begin,...,end,...]
The start position of L and R pointers
[...,begin,next,...,end,...]
R L
while L < end
1. If a user[pivot] > user[L] then move R by one and swap user[R] with user[L]
2. move L by one
After while swap user[R] with user[pivot]
Quick sort - Use the partition algorithm until every next part of the split by a pivot will have begin index greater or equals than end index.
def qsort(user, begin, end):
if begin >= end:
return
# partition
# pivot = begin
L = begin+1
R = begin
while L < end:
if user[begin] > user[L]:
R+=1
user[R], user[L] = user[L], user[R]
L+= 1
user[R], user[begin] = user[begin], user[R]
qsort(user, 0, R)
qsort(user, R+1, end)
tests = [
{'sample':[1],'answer':[1]},
{'sample':[3,9],'answer':[3,9]},
{'sample':[1,8,1],'answer':[1,1,8]},
{'sample':[7,5,5,1],'answer':[1,5,5,7]},
{'sample':[4,10,5,9,3],'answer':[3,4,5,9,10]},
{'sample':[6,6,3,8,7,7],'answer':[3,6,6,7,7,8]},
{'sample':[3,6,7,2,4,5,4],'answer':[2,3,4,4,5,6,7]},
{'sample':[1,5,6,1,9,0,7,4],'answer':[0,1,1,4,5,6,7,9]},
{'sample':[0,9,5,2,2,5,8,3,8],'answer':[0,2,2,3,5,5,8,8,9]},
{'sample':[2,5,3,3,2,0,9,0,0,7],'answer':[0,0,0,2,2,3,3,5,7,9]}
]
for test in tests:
sample = test['sample'][:]
answer = test['answer']
qsort(sample,0,len(sample))
print(sample == answer)
Solution 12:[12]
I think both answers here works ok for the list provided (which answer the original question), but would breaks if an array containing non unique values is passed. So for completeness, I would just point out the small error in each and explain how to fix them.
For example trying to sort the following array [12,4,5,6,7,3,1,15,1] (Note that 1 appears twice) with Brionius algorithm .. at some point will end up with the less array empty and the equal array with a pair of values (1,1) that can not be separated in the next iteration and the len() > 1...hence you'll end up with an infinite loop
You can fix it by either returning array if less is empty or better by not calling sort in your equal array, as in zangw answer
def sort(array=[12,4,5,6,7,3,1,15]):
less = []
equal = []
greater = []
if len(array) > 1:
pivot = array[0]
for x in array:
if x < pivot:
less.append(x)
elif x == pivot:
equal.append(x)
else: # if x > pivot
greater.append(x)
# Don't forget to return something!
return sort(less) + equal + sort(greater) # Just use the + operator to join lists
# Note that you want equal ^^^^^ not pivot
else: # You need to hande the part at the end of the recursion - when you only have one element in your array, just return the array.
return array
The fancier solution also breaks, but for a different cause, it is missing the return clause in the recursion line, which will cause at some point to return None and try to append it to a list ....
To fix it just add a return to that line
def qsort(arr):
if len(arr) <= 1:
return arr
else:
return qsort([x for x in arr[1:] if x<arr[0]]) + [arr[0]] + qsort([x for x in arr[1:] if x>=arr[0]])
Solution 13:[13]
Or if you prefer a one-liner that also illustrates the Python equivalent of C/C++ varags, lambda expressions, and if expressions:
qsort = lambda x=None, *xs: [] if x is None else qsort(*[a for a in xs if a<x]) + [x] + qsort(*[a for a in xs if a>=x])
Solution 14:[14]
def quick_sort(self, nums):
def helper(arr):
if len(arr) <= 1: return arr
#lwall is the index of the first element euqal to pivot
#rwall is the index of the first element greater than pivot
#so arr[lwall:rwall] is exactly the middle part equal to pivot after one round
lwall, rwall, pivot = 0, 0, 0
#choose rightmost as pivot
pivot = arr[-1]
for i, e in enumerate(arr):
if e < pivot:
#when element is less than pivot, shift the whole middle part to the right by 1
arr[i], arr[lwall] = arr[lwall], arr[i]
lwall += 1
arr[i], arr[rwall] = arr[rwall], arr[i]
rwall += 1
elif e == pivot:
#when element equals to pivot, middle part should increase by 1
arr[i], arr[rwall] = arr[rwall], arr[i]
rwall += 1
elif e > pivot: continue
return helper(arr[:lwall]) + arr[lwall:rwall] + helper(arr[rwall:])
return helper(nums)
Solution 15:[15]
I know many people have answered this question correctly and I enjoyed reading them. My answer is almost the same as zangw but I think the previous contributors did not do a good job of explaining visually how things actually work...so here is my attempt to help others that might visit this question/answers in the future about a simple solution for quicksort implementation.
How does it work ?
- We basically select the first item as our pivot from our list and then we create two sub lists.
- Our first sublist contains the items that are less than pivot
- Our second sublist contains our items that are great than or equal to pivot
- We then quick sort each of those and we combine them the first group + pivot + the second group to get the final result.
Here is an example along with visual to go with it ... (pivot)9,11,2,0
average: n log of n
worse case: n^2
The code:
def quicksort(data):
if (len(data) < 2):
return data
else:
pivot = data[0] # pivot
#starting from element 1 to the end
rest = data[1:]
low = [each for each in rest if each < pivot]
high = [each for each in rest if each >= pivot]
return quicksort(low) + [pivot] + quicksort(high)
items=[9,11,2,0] print(quicksort(items))
Solution 16:[16]
The algorithm contains two boundaries, one having elements less than the pivot (tracked by index "j") and the other having elements greater than the pivot (tracked by index "i").
In each iteration, a new element is processed by incrementing j.
Invariant:-
- all elements between pivot and i are less than the pivot, and
- all elements between i and j are greater than the pivot.
If the invariant is violated, ith and jth elements are swapped, and i is incremented.
After all elements have been processed, and everything after the pivot has been partitioned, the pivot element is swapped with the last element smaller than it.
The pivot element will now be in its correct place in the sequence. The elements before it will be less than it and the ones after it will be greater than it, and they will be unsorted.
def quicksort(sequence, low, high):
if low < high:
pivot = partition(sequence, low, high)
quicksort(sequence, low, pivot - 1)
quicksort(sequence, pivot + 1, high)
def partition(sequence, low, high):
pivot = sequence[low]
i = low + 1
for j in range(low + 1, high + 1):
if sequence[j] < pivot:
sequence[j], sequence[i] = sequence[i], sequence[j]
i += 1
sequence[i-1], sequence[low] = sequence[low], sequence[i-1]
return i - 1
def main(sequence):
quicksort(sequence, 0, len(sequence) - 1)
return sequence
if __name__ == '__main__':
sequence = [-2, 0, 32, 1, 56, 99, -4]
print(main(sequence))
Selecting a pivot
A "good" pivot will result in two sub-sequences of roughly the same size. Deterministically, a pivot element can either be selected in a naive manner or by computing the median of the sequence.
A naive implementation of selecting a pivot will be the first or last element. The worst-case runtime in this case will be when the input sequence is already sorted or reverse sorted, as one of the subsequences will be empty which will cause only one element to be removed per recursive call.
A perfectly balanced split is achieved when the pivot is the median element of the sequence. There are an equal number of elements greater than it and less than it. This approach guarantees a better overall running time, but is much more time-consuming.
A non-deterministic/random way of selecting the pivot would be to pick an element uniformly at random. This is a simple and lightweight approach that will minimize worst-case scenario and also lead to a roughly balanced split. This will also provide a balance between the naive approach and the median approach of selecting the pivot.
Solution 17:[17]
def quicksort(array):
if len(array) < 2:
return array
else:
pivot = array[0]
less = [i for i in array[1:] if i <= pivot]
greater = [i for i in array[1:] if i > pivot]
return quicksort(less) + [pivot] + quicksort(greater)
Solution 18:[18]
def quick_sort(array):
return quick_sort([x for x in array[1:] if x < array[0]]) + [array[0]] \
+ quick_sort([x for x in array[1:] if x >= array[0]]) if array else []
Solution 19:[19]
def Partition(A,p,q):
i=p
x=A[i]
for j in range(p+1,q+1):
if A[j]<=x:
i=i+1
tmp=A[j]
A[j]=A[i]
A[i]=tmp
l=A[p]
A[p]=A[i]
A[i]=l
return i
def quickSort(A,p,q):
if p<q:
r=Partition(A,p,q)
quickSort(A,p,r-1)
quickSort(A,r+1,q)
return A
Solution 20:[20]
A "true" in-place implementation [Algorithms 8.9, 8.11 from the Algorithm Design and Applications Book by Michael T. Goodrich and Roberto Tamassia]:
from random import randint
def partition (A, a, b):
p = randint(a,b)
# or mid point
# p = (a + b) / 2
piv = A[p]
# swap the pivot with the end of the array
A[p] = A[b]
A[b] = piv
i = a # left index (right movement ->)
j = b - 1 # right index (left movement <-)
while i <= j:
# move right if smaller/eq than/to piv
while A[i] <= piv and i <= j:
i += 1
# move left if greater/eq than/to piv
while A[j] >= piv and j >= i:
j -= 1
# indices stopped moving:
if i < j:
# swap
t = A[i]
A[i] = A[j]
A[j] = t
# place pivot back in the right place
# all values < pivot are to its left and
# all values > pivot are to its right
A[b] = A[i]
A[i] = piv
return i
def IpQuickSort (A, a, b):
while a < b:
p = partition(A, a, b) # p is pivot's location
#sort the smaller partition
if p - a < b - p:
IpQuickSort(A,a,p-1)
a = p + 1 # partition less than p is sorted
else:
IpQuickSort(A,p+1,b)
b = p - 1 # partition greater than p is sorted
def main():
A = [12,3,5,4,7,3,1,3]
print A
IpQuickSort(A,0,len(A)-1)
print A
if __name__ == "__main__": main()
Solution 21:[21]
The algorithm has 4 simple steps:
- Divide the array into 3 different parts: left, pivot and right, where pivot will have only one element. Let us choose this pivot element as the first element of array
- Append elements to the respective part by comparing them to pivot element. (explanation in comments)
- Recurse this algorithm till all elements in the array have been sorted
- Finally, join left+pivot+right parts
Code for the algorithm in python:
def my_sort(A):
p=A[0] #determine pivot element.
left=[] #create left array
right=[] #create right array
for i in range(1,len(A)):
#if cur elem is less than pivot, add elem in left array
if A[i]< p:
left.append(A[i])
#the recurssion will occur only if the left array is atleast half the size of original array
if len(left)>1 and len(left)>=len(A)//2:
left=my_sort(left) #recursive call
elif A[i]>p:
right.append(A[i]) #if elem is greater than pivot, append it to right array
if len(right)>1 and len(right)>=len(A)//2: # recurssion will occur only if length of right array is atleast the size of original array
right=my_sort(right)
A=left+[p]+right #append all three part of the array into one and return it
return A
my_sort([12,4,5,6,7,3,1,15])
Carry on with this algorithm recursively with the left and right parts.
Solution 22:[22]
Another quicksort implementation:
# A = Array
# s = start index
# e = end index
# p = pivot index
# g = greater than pivot boundary index
def swap(A,i1,i2):
A[i1], A[i2] = A[i2], A[i1]
def partition(A,g,p):
# O(n) - just one for loop that visits each element once
for j in range(g,p):
if A[j] <= A[p]:
swap(A,j,g)
g += 1
swap(A,p,g)
return g
def _quicksort(A,s,e):
# Base case - we are sorting an array of size 1
if s >= e:
return
# Partition current array
p = partition(A,s,e)
_quicksort(A,s,p-1) # Left side of pivot
_quicksort(A,p+1,e) # Right side of pivot
# Wrapper function for the recursive one
def quicksort(A):
_quicksort(A,0,len(A)-1)
A = [3,1,4,1,5,9,2,6,5,3,5,8,9,7,9,3,2,3,-1]
print(A)
quicksort(A)
print(A)
Solution 23:[23]
For Version Python 3.x: a functional-style using operator module, primarily to improve readability.
from operator import ge as greater, lt as lesser
def qsort(L):
if len(L) <= 1: return L
pivot = L[0]
sublist = lambda op: [*filter(lambda num: op(num, pivot), L[1:])]
return qsort(sublist(lesser))+ [pivot] + qsort(sublist(greater))
and is tested as
print (qsort([3,1,4,2,5]) == [1,2,3,4,5])
Solution 24:[24]
Here's an easy implementation:-
def quicksort(array):
if len(array) < 2:
return array
else:
pivot= array[0]
less = [i for i in array[1:] if i <= pivot]
greater = [i for i in array[1:] if i > pivot]
return quicksort(less) + [pivot] + quicksort(greater)
print(quicksort([10, 5, 2, 3]))
Solution 25:[25]
My answer is very similar to the great one from @alisianoi . However, I believe there is a slight inefficiency in his code (see my comment), which I removed. Moreover, I added more explanation and was a bit more specific about the problem of duplicate (pivot) values.
def quicksort(nums, begin=0, end=None):
# Only at the beginning end=None. In this case set to len(nums)-1
if end is None: end = len(nums) - 1
# If list part is invalid or has only 1 element, do nothing
if begin>=end: return
# Pick random pivot
pivot = nums[random.randint(begin, end)]
# Initialize left and right pointers
left, right = begin, end
while left < right:
# Find first "wrong" value from left hand side, i.e. first value >= pivot
# Find first "wrong" value from right hand side, i.e. first value <= pivot
# Note: In the LAST while loop, both left and right will point to pivot!
while nums[left] < pivot: left += 1
while nums[right] > pivot: right -= 1
# Swap the "wrong" values
if left != right:
nums[left], nums[right] = nums[right], nums[left]
# Problem: loop can get stuck if pivot value exists more than once. Simply solve with...
if nums[left] == nums[right]:
assert nums[left]==pivot
left += 1
# Now, left and right both point to a pivot value.
# All values to its left are smaller (or equal in case of duplicate pivot values)
# All values to its right are larger.
assert left == right and nums[left] == pivot
quicksort(nums, begin, left - 1)
quicksort(nums, left + 1, end)
return
Without recursion:
def quicksort(nums, ranges=None):
if ranges is None:
ranges = [[0, len(nums) - 1]]
while ranges != []:
[start, end] = ranges[0]
ranges = ranges[1:]
if start >= end:
continue
pivot = nums[randint(start, end)]
left = start
right = end
while left < right:
while nums[left] < pivot:
left += 1
while nums[right] > pivot:
right -= 1
if left != right:
nums[left], nums[right] = nums[right], nums[left]
if nums[left] == nums[right]:
left += 1
ranges = [[start, left - 1], [left + 1, end]] + ranges
Solution 26:[26]
- First we declare the first value in the array to be the pivot_value and we also set the left and right marks
- We create the first while loop, this while loop is there to tell the partition process to run again if it doesn't satisfy the necessary condition
- then we apply the partition process
- after both partition processes have ran, we check to see if it satisfies the proper condition. If it does, we mark it as done, if not we switch the left and right values and apply it again
- Once its done switch the left and right values and return the split_point
I am attaching the code below! This quicksort is a great learning tool because of the Location of the pivot value. Since it is in a constant place, you can walk through it multiple times and really get a hang of how it all works. In practice it is best to randomize the pivot to avoid O(N^2) runtime.
def quicksort10(alist):
quicksort_helper10(alist, 0, len(alist)-1)
def quicksort_helper10(alist, first, last):
""" """
if first < last:
split_point = partition10(alist, first, last)
quicksort_helper10(alist, first, split_point - 1)
quicksort_helper10(alist, split_point + 1, last)
def partition10(alist, first, last):
done = False
pivot_value = alist[first]
leftmark = first + 1
rightmark = last
while not done:
while leftmark <= rightmark and alist[leftmark] <= pivot_value:
leftmark = leftmark + 1
while leftmark <= rightmark and alist[rightmark] >= pivot_value:
rightmark = rightmark - 1
if leftmark > rightmark:
done = True
else:
temp = alist[leftmark]
alist[leftmark] = alist[rightmark]
alist[rightmark] = temp
temp = alist[first]
alist[first] = alist[rightmark]
alist[rightmark] = temp
return rightmark
Solution 27:[27]
def quick_sort(l):
if len(l) == 0:
return l
pivot = l[0]
pivots = [x for x in l if x == pivot]
smaller = quick_sort([x for x in l if x < pivot])
larger = quick_sort([x for x in l if x > pivot])
return smaller + pivots + larger
Solution 28:[28]
Full example with printed variables at partition step:
def partition(data, p, right):
print("\n==> Enter partition: p={}, right={}".format(p, right))
pivot = data[right]
print("pivot = data[{}] = {}".format(right, pivot))
i = p - 1 # this is a dangerous line
for j in range(p, right):
print("j: {}".format(j))
if data[j] <= pivot:
i = i + 1
print("new i: {}".format(i))
print("swap: {} <-> {}".format(data[i], data[j]))
data[i], data[j] = data[j], data[i]
print("swap2: {} <-> {}".format(data[i + 1], data[right]))
data[i + 1], data[right] = data[right], data[i + 1]
return i + 1
def quick_sort(data, left, right):
if left < right:
pivot = partition(data, left, right)
quick_sort(data, left, pivot - 1)
quick_sort(data, pivot + 1, right)
data = [2, 8, 7, 1, 3, 5, 6, 4]
print("Input array: {}".format(data))
quick_sort(data, 0, len(data) - 1)
print("Output array: {}".format(data))
Solution 29:[29]
def is_sorted(arr): #check if array is sorted
for i in range(len(arr) - 2):
if arr[i] > arr[i + 1]:
return False
return True
def qsort_in_place(arr, left, right): #arr - given array, #left - first element index, #right - last element index
if right - left < 1: #if we have empty or one element array - nothing to do
return
else:
left_point = left #set left pointer that points on element that is candidate to swap with element under right pointer or pivot element
right_point = right - 1 #set right pointer that is candidate to swap with element under left pointer
while left_point <= right_point: #while we have not checked all elements in the given array
swap_left = arr[left_point] >= arr[right] #True if we have to move that element after pivot
swap_right = arr[right_point] < arr[right] #True if we have to move that element before pivot
if swap_left and swap_right: #if both True we can swap elements under left and right pointers
arr[right_point], arr[left_point] = arr[left_point], arr[right_point]
left_point += 1
right_point -= 1
else: #if only one True we don`t have place for to swap it
if not swap_left: #if we dont need to swap it we move to next element
left_point += 1
if not swap_right: #if we dont need to swap it we move to prev element
right_point -= 1
arr[left_point], arr[right] = arr[right], arr[left_point] #swap left element with pivot
qsort_in_place(arr, left, left_point - 1) #execute qsort for left part of array (elements less than pivot)
qsort_in_place(arr, left_point + 1, right) #execute qsort for right part of array (elements most than pivot)
def main():
import random
arr = random.sample(range(1, 4000), 10) #generate random array
print(arr)
print(is_sorted(arr))
qsort_in_place(arr, 0, len(arr) - 1)
print(arr)
print(is_sorted(arr))
if __name__ == "__main__":
main()
Solution 30:[30]
This algorithm doesn't use recursive functions.
Let N be any list of numbers with len(N) > 0. Set K = [N] and execute the following program.
Note: This is a stable sorting algorithm.
def BinaryRip2Singletons(K, S):
K_L = []
K_P = [ [K[0][0]] ]
K_R = []
for i in range(1, len(K[0])):
if K[0][i] < K[0][0]:
K_L.append(K[0][i])
elif K[0][i] > K[0][0]:
K_R.append(K[0][i])
else:
K_P.append( [K[0][i]] )
K_new = [K_L]*bool(len(K_L)) + K_P + [K_R]*bool(len(K_R)) + K[1:]
while len(K_new) > 0:
if len(K_new[0]) == 1:
S.append(K_new[0][0])
K_new = K_new[1:]
else:
break
return K_new, S
N = [16, 19, 11, 15, 16, 10, 12, 14, 4, 10, 5, 2, 3, 4, 7, 1]
K = [ N ]
S = []
print('K =', K, 'S =', S)
while len(K) > 0:
K, S = BinaryRip2Singletons(K, S)
print('K =', K, 'S =', S)
PROGRAM OUTPUT:
K = [[16, 19, 11, 15, 16, 10, 12, 14, 4, 10, 5, 2, 3, 4, 7, 1]] S = []
K = [[11, 15, 10, 12, 14, 4, 10, 5, 2, 3, 4, 7, 1], [16], [16], [19]] S = []
K = [[10, 4, 10, 5, 2, 3, 4, 7, 1], [11], [15, 12, 14], [16], [16], [19]] S = []
K = [[4, 5, 2, 3, 4, 7, 1], [10], [10], [11], [15, 12, 14], [16], [16], [19]] S = []
K = [[2, 3, 1], [4], [4], [5, 7], [10], [10], [11], [15, 12, 14], [16], [16], [19]] S = []
K = [[5, 7], [10], [10], [11], [15, 12, 14], [16], [16], [19]] S = [1, 2, 3, 4, 4]
K = [[15, 12, 14], [16], [16], [19]] S = [1, 2, 3, 4, 4, 5, 7, 10, 10, 11]
K = [[12, 14], [15], [16], [16], [19]] S = [1, 2, 3, 4, 4, 5, 7, 10, 10, 11]
K = [] S = [1, 2, 3, 4, 4, 5, 7, 10, 10, 11, 12, 14, 15, 16, 16, 19]
Sources
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Source: Stack Overflow