If either i or j reach the end of their list range, how i copy the remainder of the other
list to the merge list
https://ibb.co/m9JzBYp
go to link if not get the question
def list_merge (x, y):
merge = []
i = 0
j = 0
total = len (x) + len(y)
while != total :
if x[i] < y[j]:
merge.append(x[i])
i += 1
if i >= len (x):
#how i copy the reminder
else :
merge.append(y[j])
j += 1
if j >= len (y):
#how i copy the reminder
return merge
EDIT - OP wanted the code in some specific way.. Please see second snippet.
Don't try to complicate your code.. just go with how you would do it manually and write the code.
def list_merge (x, y):
merged_list = []
i = 0
j = 0
# when one of them is empty, break out
while i < len(x) and j < len(y):
if x[i] <= y[j]:
merged_list.append(x[i])
i +=1
else:
merged_list.append(y[j])
j +=1
# if you are here, that means either one of the list is done
# so check the bounds of both lists and append the one which is not traversed
# x has some elements to add
# check how extend works... makes your code clean
if i != len(x):
merged_list.extend(x[i:])
else:
merged_list.extend(y[j:])
return merged_list
a = [1,3,5,7,10]
b = [2,4,6,8,100]
print(list_merge(a,b))
Output
[1, 2, 3, 4, 5, 6, 7, 8, 10, 100]
What OP needed
def list_merge (x, y):
merge = []
i = 0
j = 0
total = len (x) + len(y)
while len(merge) != total :
if x[i] < y[j]:
merge.append(x[i])
i += 1
if i >= len (x):
merge.extend(y[j:])
else:
merge.append(y[j])
j += 1
if j >= len (y):
merge.extend(x[i:])
return merge
a quite simple form:
def merge(*lists_in):
list_in = []
for l in lists_in:
list_in += l
i = 0
while True:
if i == list_in.__len__() -1:
break
if list_in[i] > list_in[i+1]:
temp = list_in[i]
list_in[i] = list_in[i+1]
list_in[i+1] = temp
i = 0
else:
i += 1
return list_in
Testing it:
list1 = [1,4]
list2 = [1,5,6]
list3 = [3,7,9,10]
print(merge(list1, list2, list3))
Out:
[1, 1, 3, 4, 5, 6, 7, 9, 10]
This can be solved rather nicely using deques, which allow you to efficiently look at and remove the first element.
from collections import deque
# A generator function that interleaves two sorted deques
# into a single sorted sequence.
def merge_deques(x, y):
while x and y:
yield (x if x[0] <= y[0] else y).popleft()
# When we reach this point, one of x or y is empty,
# so one of these doesn't yield any values.
yield from x
yield from y
# Makes a list by consuming the merged sequence of two deques.
def list_merge(x, y):
return list(merge_deques(deque(x), deque(y))
I have the task to print all subsets of given length, I have created the functions to print out the all the subsets. Everything works fine the subsets are generated, but the output is wrong for each call. For example if I call print(get_subsets([1,2,3],2)) the output is [1,2] [1,3] [2,3] and [3]. Of course 3 is not supposed to be there and I can't figure out why. Any help will be appreciated and feedback of course.
def get_subsets(nums, k):
all_subsets = []
_gen_subsets(nums =nums,curr_idx =0,curr_subset=[],
all_subsets=all_subsets)
for curr_subset in all_subsets:
if len(curr_subset) > k or len(curr_subset) < k:
all_subsets.remove(curr_subset)
return all_subsets
def _gen_subsets(nums,curr_idx, curr_subset, all_subsets):
if curr_idx >= len(nums):
all_subsets.append(curr_subset)
else:
itr_subset = curr_subset.copy()
itr_subset.append(nums[curr_idx])
_gen_subsets(nums=nums,
curr_idx=curr_idx+1,
curr_subset=itr_subset,
all_subsets=all_subsets)
_gen_subsets(nums=nums,
curr_idx=curr_idx+1,
curr_subset=curr_subset,
all_subsets=all_subsets)
You are trying to iterate through a list from which you are also removing elements. This messed up the position and index of the list of subsets and gave you the wrong output.
Change the function get_subsets to this:
def get_subsets(nums, k):
all_subsets = []
_gen_subsets(nums =nums,curr_idx =0,curr_subset=[],
all_subsets=all_subsets)
final_subset = all_subsets.copy()
for n in all_subsets:
if len(n) != k:
final_subset.remove(n)
return final_subset
Or this:
def get_subsets(nums, k):
all_subsets = []
_gen_subsets(nums =nums,curr_idx =0,curr_subset=[],
all_subsets=all_subsets)
all_subsets = [n for n in all_subsets if len(n) == k]
return all_subsets
Your problem is caused by this:
for curr_subset in all_subsets:
if len(curr_subset) > k or len(curr_subset) < k:
all_subsets.remove(curr_subset)
You are removing array elements while iterating. That leads to the following scenario: Initially you have array [X1, X2, X3].
In the first iteration, you remove the first element X1. Then the second element X2 becomes the first element and therefore, on the second iteration, the third element X3 (which has become the second element) is traversed and X2 is skipped. That is why [3] is not removed from your array.
To solve this, you can change get_subsets function as below:
def get_subsets(nums, k):
all_subsets = []
_gen_subsets(nums =nums,curr_idx =0,curr_subset=[],
all_subsets=all_subsets)
all_subsets = [subset for subset in all_subsets if len(subset) == k]
return all_subsets
maybe thes two will help you?
def Combinations(l, d):
if (d <= 1) or (len(l) <= 1):
return [[i] for i in l]
else:
result = []
for i in range(len(l)):
c = l[i]
lx = l[:i]+l[i+1:]
for cmb in Combinations(lx, d-1):
result.append([c]+cmb)
return result
print(*Combinations([1, 2, 3, 4, 5, 6], 2), sep='\n')
def Allocations(l, d):
if (d <= 1) or (len(l) <= 1):
return [[i] for i in l]
else:
result = []
for i in range(len(l)+1-d):
c = l[i]
lx = l[i+1:]
for cmb in Allocations(lx, d-1):
result.append([c]+cmb)
return result
print(*Allocations([1, 2, 3, 4, 5, 6], 3), sep='\n')
I want to find a sequence of n consecutive integers within a sorted list and return that sequence. This is the best I can figure out (for n = 4), and it doesn't allow the user to specify an n.
my_list = [2,3,4,5,7,9]
for i in range(len(my_list)):
if my_list[i+1] == my_list[i]+1 and my_list[i+2] == my_list[i]+2 and my_list[i+3] == my_list[i]+3:
my_sequence = list(range(my_list[i],my_list[i]+4))
my_sequence = [2,3,4,5]
I just realized this code doesn't work and returns an "index out of range" error, so I'll have to mess with the range of the for loop.
Here's a straight-forward solution. It's not as efficient as it might be, but it will be fine unless you have very long lists:
myarray = [2,5,1,7,3,8,1,2,3,4,5,7,4,9,1,2,3,5]
for idx, a in enumerate(myarray):
if myarray[idx:idx+4] == [a,a+1,a+2,a+3]:
print([a, a+1,a+2,a+3])
break
Create a nested master result list, then go through my_sorted_list and add each item to either the last list in the master (if discontinuous) or to a new list in the master (if continuous):
>>> my_sorted_list = [0,2,5,7,8,9]
>>> my_sequences = []
>>> for idx,item in enumerate(my_sorted_list):
... if not idx or item-1 != my_sequences[-1][-1]:
... my_sequences.append([item])
... else:
... my_sequences[-1].append(item)
...
>>> max(my_sequences, key=len)
[7, 8, 9]
A short and concise way is to fill an array with numbers every time you find the next integer is the current integer plus 1 (until you already have N consecutive numbers in array), and for anything else, we can empty the array:
arr = [4,3,1,2,3,4,5,7,5,3,2,4]
N = 4
newarr = []
for i in range(len(arr)-1):
if(arr[i]+1 == arr[i+1]):
newarr += [arr[i]]
if(len(newarr) == N):
break
else:
newarr = []
When the code is run, newarr will be:
[1, 2, 3, 4]
#size = length of sequence
#span = the span of neighbour integers
#the time complexity is O(n)
def extractSeq(lst,size,span=1):
lst_size = len(lst)
if lst_size < size:
return []
for i in range(lst_size - size + 1):
for j in range(size - 1):
if lst[i + j] + span == lst[i + j + 1]:
continue
else:
i += j
break
else:
return lst[i:i+size]
return []
mylist = [2,3,4,5,7,9]
for j in range(len(mylist)):
m=mylist[j]
idx=j
c=j
for i in range(j,len(mylist)):
if mylist[i]<m:
m=mylist[i]
idx=c
c+=1
tmp=mylist[j]
mylist[j]=m
mylist[idx]=tmp
print(mylist)
I tried implementing Insertion sort with for loops only and wrote the following code:
def isort(L): #implementation with a for loop
for i in range(1,len(L)):
small = L[i]
M = range(i)
M.reverse()
for j in M:
if small<L[j]:
L[j+1]=L[j]
else:
break
L[j+1] = small
return L
L = [5,4,3,2,1]
M = isort(L)
print M
This gives the output [5,1,2,3,4]. Can someone please point out where I am making a mistake
Change (the fix shown in the question is easy, the one-off error was caused by one little +1 :)):
L[j+1] = small
To:
L[j] = small
Testing:
>>> isort([5, 4, 3, 2, 1])
[1, 2, 3, 4, 5]
However, there are some other things with your code, as illustrated- it will not work alot of the time. With a fair few tweaks, we can get it to work:
def isort(L):
for i in range(1,len(L)):
small = L[i]
M = range(-1, i)
M.reverse()
for j in M:
if j>=0 and small<L[j]:
L[j+1]=L[j]
else:
break
L[j+1] = small
return L
Testing:
>>> isort([4, 5, 3, 2, 1])
[1, 2, 3, 4, 5]
The post condition for the inner loop is that j is pointing for the first value that is smaller than small (this is achieved by the break call). However, the loop naturally exists when j=0, therefore in every last inner iteration, the condition is not what you'd expect.
To fix it, I suggest initializing M from -1:
M = range(-1, i)
But then, you have to check as well that j is positive (to avoid making changes you don't want to):
if j>=0 and small<L[j]:
L[j+1]=L[j]
This is little tricky :
I took the inner loop range function as range(j, -2, -1) , so the inner loop always breaks at one position ahead, so the last statement arr[j + 1] = key works perfectly
arr = [5, 4, 3, 2, 1]
for i in range(1, len(arr)):
j = i - 1
key = arr[i]
for j in range(j, -2, -1):
if j < 0 or key >= arr[j]:
break
else:
arr[j + 1] = arr[j]
arr[j + 1] = key
if __name__ == "__main__":
n = int(input("How many numbers ?\t"))
nums = [int(x) for x in input("Enter {} numbers\t".format(n)).split()]
for i in range(1,n):
val = nums[i]
for j in range(i-1,-2,-1):
if j < 0 : break
if nums[j] > val:
nums[j+1] = nums[j]
else:
break
nums[j+1] = val
for num in nums:
print(num,end=' ')
print()
In class we are doing sorting algorithms and, although I understand them fine when talking about them and writing pseudocode, I am having problems writing actual code for them.
This is my attempt in Python:
mylist = [12, 5, 13, 8, 9, 65]
def bubble(badList):
length = len(badList) - 1
unsorted = True
while unsorted:
for element in range(0,length):
unsorted = False
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
print badList
else:
unsorted = True
print bubble(mylist)
Now, this (as far as I can tell) sorts correctly, but once it finishes it just loops indefinitely.
How can this code be fixed so the function finishes properly and correctly sorts a list of any (reasonable) size?
P.S. I know I should not really have prints in a function and I should have a return, but I just have not done that yet as my code does not really work yet.
To explain why your script isn't working right now, I'll rename the variable unsorted to sorted.
At first, your list isn't yet sorted. Of course, we set sorted to False.
As soon as we start the while loop, we assume that the list is already sorted. The idea is this: as soon as we find two elements that are not in the right order, we set sorted back to False. sorted will remain True only if there were no elements in the wrong order.
sorted = False # We haven't started sorting yet
while not sorted:
sorted = True # Assume the list is now sorted
for element in range(0, length):
if badList[element] > badList[element + 1]:
sorted = False # We found two elements in the wrong order
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
# We went through the whole list. At this point, if there were no elements
# in the wrong order, sorted is still True. Otherwise, it's false, and the
# while loop executes again.
There are also minor little issues that would help the code be more efficient or readable.
In the for loop, you use the variable element. Technically, element is not an element; it's a number representing a list index. Also, it's quite long. In these cases, just use a temporary variable name, like i for "index".
for i in range(0, length):
The range command can also take just one argument (named stop). In that case, you get a list of all the integers from 0 to that argument.
for i in range(length):
The Python Style Guide recommends that variables be named in lowercase with underscores. This is a very minor nitpick for a little script like this; it's more to get you accustomed to what Python code most often resembles.
def bubble(bad_list):
To swap the values of two variables, write them as a tuple assignment. The right hand side gets evaluated as a tuple (say, (badList[i+1], badList[i]) is (3, 5)) and then gets assigned to the two variables on the left hand side ((badList[i], badList[i+1])).
bad_list[i], bad_list[i+1] = bad_list[i+1], bad_list[i]
Put it all together, and you get this:
my_list = [12, 5, 13, 8, 9, 65]
def bubble(bad_list):
length = len(bad_list) - 1
sorted = False
while not sorted:
sorted = True
for i in range(length):
if bad_list[i] > bad_list[i+1]:
sorted = False
bad_list[i], bad_list[i+1] = bad_list[i+1], bad_list[i]
bubble(my_list)
print my_list
(I removed your print statement too, by the way.)
The goal of bubble sort is to move the heavier items at the bottom in each round, while moving the lighter items up. In the inner loop, where you compare the elements, you don't have to iterate the whole list in each turn. The heaviest is already placed last. The swapped variable is an extra check so we can mark that the list is now sorted and avoid continuing with unnecessary calculations.
def bubble(badList):
length = len(badList)
for i in range(0,length):
swapped = False
for element in range(0, length-i-1):
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
swapped = True
if not swapped: break
return badList
Your version 1, corrected:
def bubble(badList):
length = len(badList) - 1
unsorted = True
while unsorted:
unsorted = False
for element in range(0,length):
#unsorted = False
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
unsorted = True
#print badList
#else:
#unsorted = True
return badList
This is what happens when you use variable name of negative meaning, you need to invert their values. The following would be easier to understand:
sorted = False
while not sorted:
...
On the other hand, the logic of the algorithm is a little bit off. You need to check whether two elements swapped during the for loop. Here's how I would write it:
def bubble(values):
length = len(values) - 1
sorted = False
while not sorted:
sorted = True
for element in range(0,length):
if values[element] > values[element + 1]:
hold = values[element + 1]
values[element + 1] = values[element]
values[element] = hold
sorted = False
return values
Your use of the Unsorted variable is wrong; you want to have a variable that tells you if you have swapped two elements; if you have done that, you can exit your loop, otherwise, you need to loop again. To fix what you've got here, just put the "unsorted = false" in the body of your if case; remove your else case; and put "unsorted = true before your for loop.
def bubble_sort(l):
for passes_left in range(len(l)-1, 0, -1):
for index in range(passes_left):
if l[index] < l[index + 1]:
l[index], l[index + 1] = l[index + 1], l[index]
return l
#A very simple function, can be optimized (obviously) by decreasing the problem space of the 2nd array. But same O(n^2) complexity.
def bubble(arr):
l = len(arr)
for a in range(l):
for b in range(l-1):
if (arr[a] < arr[b]):
arr[a], arr[b] = arr[b], arr[a]
return arr
You've got a couple of errors in there. The first is in length, and the second is in your use of unsorted (as stated by McWafflestix). You probably also want to return the list if you're going to print it:
mylist = [12, 5, 13, 8, 9, 65]
def bubble(badList):
length = len(badList) - 2
unsorted = True
while unsorted:
for element in range(0,length):
unsorted = False
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
print badList
unsorted = True
return badList
print bubble(mylist)
eta: You're right, the above is buggy as hell. My bad for not testing through some more examples.
def bubble2(badList):
swapped = True
length = len(badList) - 2
while swapped:
swapped = False
for i in range(0, length):
if badList[i] > badList[i + 1]:
# swap
hold = badList[i + 1]
badList[i + 1] = badList[i]
badList[i] = hold
swapped = True
return badList
I am a fresh fresh beginner, started to read about Python yesterday.
Inspired by your example I created something maybe more in the 80-ties style, but nevertheless it kinda works
lista1 = [12, 5, 13, 8, 9, 65]
i=0
while i < len(lista1)-1:
if lista1[i] > lista1[i+1]:
x = lista1[i]
lista1[i] = lista1[i+1]
lista1[i+1] = x
i=0
continue
else:
i+=1
print(lista1)
The problem with the original algorithm is that if you had a lower number further in the list, it would not bring it to the correct sorted position. The program needs to go back the the beginning each time to ensure that the numbers sort all the way through.
I simplified the code and it will now work for any list of numbers regardless of the list and even if there are repeating numbers. Here's the code
mylist = [9, 8, 5, 4, 12, 1, 7, 5, 2]
print mylist
def bubble(badList):
length = len(badList) - 1
element = 0
while element < length:
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
element = 0
print badList
else:
element = element + 1
print bubble(mylist)
def bubble_sort(l):
exchanged = True
iteration = 0
n = len(l)
while(exchanged):
iteration += 1
exchanged = False
# Move the largest element to the end of the list
for i in range(n-1):
if l[i] > l[i+1]:
exchanged = True
l[i], l[i+1] = l[i+1], l[i]
n -= 1 # Largest element already towards the end
print 'Iterations: %s' %(iteration)
return l
def bubbleSort(alist):
if len(alist) <= 1:
return alist
for i in range(0,len(alist)):
print "i is :%d",i
for j in range(0,i):
print "j is:%d",j
print "alist[i] is :%d, alist[j] is :%d"%(alist[i],alist[j])
if alist[i] > alist[j]:
alist[i],alist[j] = alist[j],alist[i]
return alist
alist = [54,26,93,17,77,31,44,55,20,-23,-34,16,11,11,11]
print bubbleSort(alist)
def bubble_sort(a):
t = 0
sorted = False # sorted = False because we have not began to sort
while not sorted:
sorted = True # Assume sorted = True first, it will switch only there is any change
for key in range(1,len(a)):
if a[key-1] > a[key]:
sorted = False
t = a[key-1]; a[key-1] = a[key]; a[key] = t;
print a
A simpler example:
a = len(alist)-1
while a > 0:
for b in range(0,a):
#compare with the adjacent element
if alist[b]>=alist[b+1]:
#swap both elements
alist[b], alist[b+1] = alist[b+1], alist[b]
a-=1
This simply takes the elements from 0 to a(basically, all the unsorted elements in that round) and compares it with its adjacent element, and making a swap if it is greater than its adjacent element. At the end the round, the last element is sorted, and the process runs again without it, until all elements have been sorted.
There is no need for a condition whether sort is true or not.
Note that this algorithm takes into consideration the position of the numbers only when swapping, so repeated numbers will not affect it.
PS. I know it has been very long since this question was posted, but I just wanted to share this idea.
def bubble_sort(li):
l = len(li)
tmp = None
sorted_l = sorted(li)
while (li != sorted_l):
for ele in range(0,l-1):
if li[ele] > li[ele+1]:
tmp = li[ele+1]
li[ele+1] = li [ele]
li[ele] = tmp
return li
def bubbleSort ( arr ):
swapped = True
length = len ( arr )
j = 0
while swapped:
swapped = False
j += 1
for i in range ( length - j ):
if arr [ i ] > arr [ i + 1 ]:
# swap
tmp = arr [ i ]
arr [ i ] = arr [ i + 1]
arr [ i + 1 ] = tmp
swapped = True
if __name__ == '__main__':
# test list
a = [ 67, 45, 39, -1, -5, -44 ];
print ( a )
bubbleSort ( a )
print ( a )
def bubblesort(array):
for i in range(len(array)-1):
for j in range(len(array)-1-i):
if array[j] > array[j+1]:
array[j], array[j+1] = array[j+1], array[j]
return(array)
print(bubblesort([3,1,6,2,5,4]))
arr = [5,4,3,1,6,8,10,9] # array not sorted
for i in range(len(arr)):
for j in range(i, len(arr)):
if(arr[i] > arr[j]):
arr[i], arr[j] = arr[j], arr[i]
print (arr)
I consider adding my solution because ever solution here is having
greater time
greater space complexity
or doing too much operations
then is should be
So, here is my solution:
def countInversions(arr):
count = 0
n = len(arr)
for i in range(n):
_count = count
for j in range(0, n - i - 1):
if arr[j] > arr[j + 1]:
count += 1
arr[j], arr[j + 1] = arr[j + 1], arr[j]
if _count == count:
break
return count
If anyone is interested in a shorter implementation using a list comprehension:
def bubble_sort(lst: list) -> None:
[swap_items(lst, i, i+1) for left in range(len(lst)-1, 0, -1) for i in range(left) if lst[i] > lst[i+1]]
def swap_items(lst: list, pos1: int, pos2: int) -> None:
lst[pos1], lst[pos2] = lst[pos2], lst[pos1]
Here is a different variation of bubble sort without for loop. Basically you are considering the lastIndex of the array and slowly decrementing it until it first index of the array.
The algorithm will continue to move through the array like this until an entire pass is made without any swaps occurring.
The bubble is sort is basically Quadratic Time: O(n²) when it comes to performance.
class BubbleSort:
def __init__(self, arr):
self.arr = arr;
def bubbleSort(self):
count = 0;
lastIndex = len(self.arr) - 1;
while(count < lastIndex):
if(self.arr[count] > self.arr[count + 1]):
self.swap(count)
count = count + 1;
if(count == lastIndex):
count = 0;
lastIndex = lastIndex - 1;
def swap(self, count):
temp = self.arr[count];
self.arr[count] = self.arr[count + 1];
self.arr[count + 1] = temp;
arr = [9, 1, 5, 3, 8, 2]
p1 = BubbleSort(arr)
print(p1.bubbleSort())
def bubblesort(L,s):
if s >-1 :
bubblesort(L,s-1)
for i in range(len(L)-1-s):
if L[i]>L[i+1]:
temp = L[i+1]
L[i+1] = L[i]
L[i] = temp
return L
Nlist = [3,50,7,1,8,11,9,0,-1,5]
print(bubblesort(Nlist,len(Nlist)))
Answers provided by the-fury and Martin Cote fixed the problem of the infinite loop, but my code would still not work correctly (for a larger list, it would not sort correctly.). I ended up ditching the unsorted variable and used a counter instead.
def bubble(badList):
length = len(badList) - 1
n = 0
while n < len(badList):
for element in range(0,length):
if badList[element] > badList[element + 1]:
hold = badList[element + 1]
badList[element + 1] = badList[element]
badList[element] = hold
n = 0
else:
n += 1
return badList
if __name__ == '__main__':
mylist = [90, 10, 2, 76, 17, 66, 57, 23, 57, 99]
print bubble(mylist)
If anyone could provide any pointers on how to improve my code in the comments, it would be much appreciated.
Try this
a = int(input("Enter Limit"))
val = []
for z in range(0,a):
b = int(input("Enter Number in List"))
val.append(b)
for y in range(0,len(val)):
for x in range(0,len(val)-1):
if val[x]>val[x+1]:
t = val[x]
val[x] = val[x+1]
val[x+1] = t
print(val)
idk if this might help you after 9 years...
its a simple bubble sort program
l=[1,6,3,7,5,9,8,2,4,10]
for i in range(1,len(l)):
for j in range (i+1,len(l)):
if l[i]>l[j]:
l[i],l[j]=l[j],l[i]
def merge_bubble(arr):
k = len(arr)
while k>2:
for i in range(0,k-1):
for j in range(0,k-1):
if arr[j] > arr[j+1]:
arr[j],arr[j+1] = arr[j+1],arr[j]
return arr
break
else:
if arr[0] > arr[1]:
arr[0],arr[1] = arr[1],arr[0]
return arr
def bubble_sort(l):
for i in range(len(l) -1):
for j in range(len(l)-i-1):
if l[j] > l[j+1]:
l[j],l[j+1] = l[j+1], l[j]
return l
def bubble_sorted(arr:list):
while True:
for i in range(0,len(arr)-1):
count = 0
if arr[i] > arr[i+1]:
count += 1
arr[i], arr[i+1] = arr[i+1], arr[i]
if count == 0:
break
return arr
arr = [30,20,80,40,50,10,60,70,90]
print(bubble_sorted(arr))
#[20, 30, 40, 50, 10, 60, 70, 80, 90]
def bubbleSort(a):
def swap(x, y):
temp = a[x]
a[x] = a[y]
a[y] = temp
#outer loop
for j in range(len(a)):
#slicing to the center, inner loop, python style
for i in range(j, len(a) - j):
#find the min index and swap
if a[i] < a[j]:
swap(j, i)
#find the max index and swap
if a[i] > a[len(a) - j - 1]:
swap(len(a) - j - 1, i)
return a