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Code for consecutive strings works but can't pass random tests

In this problem, I'm given an array(list) strarr of strings and an integer k. My task is to return the first longest string consisting of k consecutive strings taken in the array. My code passed all the sample tests from CodeWars but can't seem to pass the random tests.
Here's the link to the problem.
I did it in two days. I found the max consecutively combined string first. Here's the code for that.
strarr = []
def longest_consec(strarr, k):
strarr.append('')
length = len(strarr)
cons_list = []
end = k
start = 0
freq = -length/2
final_string = []
largest = max(strarr, key=len, default='')
if k == 1:
return largest
elif 1 < k < length:
while(freq <= 1):
cons_list.append(strarr[start:end])
start += k-1
end += k-1
freq += 1
for index in cons_list:
final_string.append(''.join(index))
return max(final_string, key=len, default='')
else:
return ""
Since that didn't pass all the random tests, I compared the combined k strings on both sides of the single largest string. But, this way, the code doesn't account for the case when the single largest string is in the middle. Please help.
strarr = []
def longest_consec(strarr, k):
strarr.append('')
length = len(strarr)
largest = max(strarr, key=len, default='')
pos = int(strarr.index(largest))
if k == 1:
return largest
elif 1 < k < length:
prev_string = ''.join(strarr[pos+1-k:pos+1])
next_string = ''.join(strarr[pos:pos+k])
if len(prev_string) >= len(next_string):
res = prev_string
else:
res = next_string
return res
else:
return ""
print(longest_consec(["zone", "abigail", "theta", "form", "libe"], 2))

Let's start from the first statement of your function:
if k == 1:
while(p <= 1):
b.append(strarr[j:i])
j += 1
i += 1
p += 1
for w in b:
q.append(''.join(w))
return max(q, key=len)
Here q is finally equal strarr so you can shorten this code to:
if k == 1:
return max(strarr, key=len)
I see that second statement's condition checks if k value is between 1 and length of string array inclusive:
elif k > 1 and k <= 2*a:
...
If you want no errors remove equality symbol, last element of every array has index lesser than its length (equal exactly length of it minus 1).
Ceiling and division is not necessary in a definition, so you can shorten this:
a = ceil(len(strarr)/2)
into this:
a = len(strarr)
then your elif statement may look like below:
elif 1 < k < a: # Same as (k > 1 and k < a)
...
again, I see you want to concatenate (add) the longest string to k next strings using this code:
while(p <= 1):
b.append(strarr[j:i])
j += k-1
i += k-1
p += 1
for w in b:
q.append(''.join(w))
return max(q, key=len)
the more clearer way of doing this:
longest = max(strarr, key=len) # Longest string in array.
index = 0 # Index of the current item.
for string in strarr:
# If current string is equal the longest one ...
if string == longest:
# Join 'k' strings from current index (longest string index).
return ''.join(strarr[index:index + k])
index += 1 # Increase current index.
And the last statement which is:
elif k > 2*a or k<1:
return ""
if all previous statements failed then value is invalid so you can instead write:
return "" # Same as with else.
Now everything should work. I advice you learning the basics (especially lists, strings and slices), and please name your variables wisely so they are more readable.

You can try this as well
this has passed all the test cases on the platform you suggested.
def longest_consec(strarr, k):
i = 0
max_ = ""
res = ""
if (k<=0) or (k>len(strarr)):
return ""
while i<=(len(strarr)-k):
start = "".join(strarr[i:i+k])
max_ = max(max_, start, key=len)
if max_==start:
res=strarr[i:i+k]
i+=1
return max_
#output: ["zone", "abigail", "theta", "form", "libe", "zas", "theta", "abigail"], 2 -> abigailtheta
#output: ["zones", "abigail", "theta", "form", "libe", "zas", "theta", "abigail"],2 -> zonesabigail

Finding first pair of numbers in array that sum to value

Im trying to solve the following Codewars problem: https://www.codewars.com/kata/sum-of-pairs/train/python
Here is my current implementation in Python:
def sum_pairs(ints, s):
right = float("inf")
n = len(ints)
m = {}
dup = {}
for i, x in enumerate(ints):
if x not in m.keys():
m[x] = i # Track first index of x using hash map.
elif x in m.keys() and x not in dup.keys():
dup[x] = i
for x in m.keys():
if s - x in m.keys():
if x == s-x and x in dup.keys():
j = m[x]
k = dup[x]
else:
j = m[x]
k = m[s-x]
comp = max(j,k)
if comp < right and j!= k:
right = comp
if right > n:
return None
return [s - ints[right],ints[right]]
The code seems to produce correct results, however the input can consist of array with up to 10 000 000 elements, so the execution times out for large inputs. I need help with optimizing/modifying the code so that it can handle sufficiently large arrays.

Your code inefficient for large list test cases so it gives timeout error. Instead you can do:
def sum_pairs(lst, s):
seen = set()
for item in lst:
if s - item in seen:
return [s - item, item]
seen.add(item)
We put the values in seen until we find a value that produces the specified sum with one of the seen values.
For more information go: Referance link

Maybe this code:
def sum_pairs(lst, s):
c = 0
while c<len(lst)-1:
if c != len(lst)-1:
x= lst[c]
spam = c+1
while spam < len(lst):
nxt= lst[spam]
if nxt + x== s:
return [x, nxt]
spam += 1
else:
return None
c +=1
lst = [5, 6, 5, 8]
s = 14
print(sum_pairs(lst, s))
Output:
[6, 8]

This answer unfortunately still times out, even though it's supposed to run in O(n^3) (since it is dominated by the sort, the rest of the algorithm running in O(n)). I'm not sure how you can obtain better than this complexity, but I thought I might put this idea out there.
def sum_pairs(ints, s):
ints_with_idx = enumerate(ints)
# Sort the array of ints
ints_with_idx = sorted(ints_with_idx, key = lambda (idx, num) : num)
diff = 1000000
l = 0
r = len(ints) - 1
# Indexes of the sum operands in sorted array
lSum = 0
rSum = 0
while l < r:
# Compute the absolute difference between the current sum and the desired sum
sum = ints_with_idx[l][1] + ints_with_idx[r][1]
absDiff = abs(sum - s)
if absDiff < diff:
# Update the best difference
lSum = l
rSum = r
diff = absDiff
elif sum > s:
# Decrease the large value
r -= 1
else:
# Test to see if the indexes are better (more to the left) for the same difference
if absDiff == diff:
rightmostIdx = max(ints_with_idx[l][0], ints_with_idx[r][0])
if rightmostIdx < max(ints_with_idx[lSum][0], ints_with_idx[rSum][0]):
lSum = l
rSum = r
# Increase the small value
l += 1
# Retrieve indexes of sum operands
aSumIdx = ints_with_idx[lSum][0]
bSumIdx = ints_with_idx[rSum][0]
# Retrieve values of operands for sum in correct order
aSum = ints[min(aSumIdx, bSumIdx)]
bSum = ints[max(aSumIdx, bSumIdx)]
if aSum + bSum == s:
return [aSum, bSum]
else:
return None

Find/extract a sequence of integers within a list in python

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)

How to return alphabetical substrings?

I'm trying to write a function that takes a string s as an input and returns a list of those substrings within s that are alphabetical. For example, s = 'acegibdh' should return ['acegi', 'bdh'].
Here's the code I've come up with:
s = 'acegibdh'
ans = []
subs = []
i = 0
while i != len(s) - 1:
while s[i] < s[i+1]:
subs.append(s[i])
i += 1
if s[i] > s[i-1]:
subs.append(s[i])
i += 1
subs = ''.join(subs)
ans.append(subs)
subs = []
print ans
It keeps having trouble with the last letter of the string, because of the i+1 test going beyond the index range. I've spent a long time tinkering with it to try and come up with a way to avoid that problem. Does anyone know how to do this?

Why not hard-code the first letter into ans, and then just work with the rest of the string? You can just iterate over the string itself instead of using indices.
>>> s = 'acegibdh'
>>> ans = []
>>> ans.append(s[0])
>>> for letter in s[1:]:
... if letter >= ans[-1][-1]:
... ans[-1] += letter
... else:
... ans.append(letter)
...
>>> ans
['acegi', 'bdh']

s = 'acegibdh'
ans = []
subs = []
subs.append(s[0])
for x in range(len(s)-1):
if s[x] <= s[x+1]:
subs.append(s[x+1])
if s[x] > s[x+1]:
subs = ''.join(subs)
ans.append(subs)
subs = []
subs.append(s[x+1])
subs = ''.join(subs)
ans.append(subs)
print ans
I decided to change your code a bit let me know if you have any questions

Just for fun, a one line solution.
>>> s='acegibdh'
>>> [s[l:r] for l,r in (lambda seq:zip(seq,seq[1:]))([0]+[idx+1 for idx in range(len(s)-1) if s[idx]>s[idx+1]]+[len(s)])]
['acegi', 'bdh']

You should try to avoid loops that increment the position by more than one char per iteration.
Often it is more clear to introduce an additional variable to store information about the previous state:
s = 'acegibdh'
prev = None
ans = []
subs = []
for ch in s:
if prev is None or ch > prev:
subs.append(ch)
else:
ans.append(''.join(subs))
subs = [ch]
prev = ch
ans.append(''.join(subs))
I think this read more straight forward (if there is no previous character or it's still alphabetical with the current char append, else start a new substring). Also you can't get index out of range problems with this approch.

More than one while loop is overkill. I think this is simpler and satisfies your requirement. Note, this fails on empty string.
s = 'acegibdh'
ans = []
current = str(s[0])
i = 1
while i < len(s):
if s[i] > s[i-1]:
current += s[i]
else:
ans.append(current)
current = ''
i += 1
if current != '':
ans.append(current)
print ans

just for fun cause I like doing things a little different sometimes
from itertools import groupby,chain,cycle
def my_gen(s):
check = cycle([1,0])
for k,v in groupby(zip(s,s[1:]),lambda x:x[0]<x[1]):
if k:
v = zip(*v)
yield v[0] + (v[1][-1],)
print list(my_gen('acegibdhabcdefghijk'))

Some of the solutions posted have an index error for empty strings.
Also, instead of keeping a list of characters, or doing repeated string concatenations, you can track the start index, i, of a solution substring and yield s[i:j] where s[j] < s[j-1], then set i to j.
Generator that yields substrings when the next letter is lexicographically less than the previous:
def alpha_subs(s):
i, j = 0, 1
while j < len(s):
if s[j] < s[j-1]:
yield s[i:j]
i = j
j += 1
if s[i:j]:
yield s[i:j]
print(list(alpha_subs('')))
print(list(alpha_subs('acegibdh')))
print(list(alpha_subs('acegibdha')))
[]
['acegi', 'bdh']
['acegi', 'bdh', 'a']
For case insensitivity:
def alpha_subs(s, ignore_case=False):
qs = s.lower() if ignore_case else s
i, j = 0, 1
while j < len(s):
if qs[j] < qs[j-1]:
yield s[i:j]
i = j
j += 1
if s[i:j]:
yield s[i:j]
print(list(alpha_subs('acEgibDh', True)))
print(list(alpha_subs('acEgibDh')))
['acEgi', 'bDh']
['ac', 'Egi', 'b', 'Dh']

Finding all possible permutations of a given string in python

I have a string. I want to generate all permutations from that string, by changing the order of characters in it. For example, say:
x='stack'
what I want is a list like this,
l=['stack','satck','sackt'.......]
Currently I am iterating on the list cast of the string, picking 2 letters randomly and transposing them to form a new string, and adding it to set cast of l. Based on the length of the string, I am calculating the number of permutations possible and continuing iterations till set size reaches the limit.
There must be a better way to do this.

The itertools module has a useful method called permutations(). The documentation says:
itertools.permutations(iterable[, r])
Return successive r length permutations of elements in the iterable.
If r is not specified or is None, then r defaults to the length of the
iterable and all possible full-length permutations are generated.
Permutations are emitted in lexicographic sort order. So, if the input
iterable is sorted, the permutation tuples will be produced in sorted
order.
You'll have to join your permuted letters as strings though.
>>> from itertools import permutations
>>> perms = [''.join(p) for p in permutations('stack')]
>>> perms
['stack', 'stakc', 'stcak', 'stcka', 'stkac', 'stkca', 'satck',
'satkc', 'sactk', 'sackt', 'saktc', 'sakct', 'sctak', 'sctka',
'scatk', 'scakt', 'sckta', 'sckat', 'sktac', 'sktca', 'skatc',
'skact', 'skcta', 'skcat', 'tsack', 'tsakc', 'tscak', 'tscka',
'tskac', 'tskca', 'tasck', 'taskc', 'tacsk', 'tacks', 'taksc',
'takcs', 'tcsak', 'tcska', 'tcask', 'tcaks', 'tcksa', 'tckas',
'tksac', 'tksca', 'tkasc', 'tkacs', 'tkcsa', 'tkcas', 'astck',
'astkc', 'asctk', 'asckt', 'asktc', 'askct', 'atsck', 'atskc',
'atcsk', 'atcks', 'atksc', 'atkcs', 'acstk', 'acskt', 'actsk',
'actks', 'ackst', 'ackts', 'akstc', 'aksct', 'aktsc', 'aktcs',
'akcst', 'akcts', 'cstak', 'cstka', 'csatk', 'csakt', 'cskta',
'cskat', 'ctsak', 'ctska', 'ctask', 'ctaks', 'ctksa', 'ctkas',
'castk', 'caskt', 'catsk', 'catks', 'cakst', 'cakts', 'cksta',
'cksat', 'cktsa', 'cktas', 'ckast', 'ckats', 'kstac', 'kstca',
'ksatc', 'ksact', 'kscta', 'kscat', 'ktsac', 'ktsca', 'ktasc',
'ktacs', 'ktcsa', 'ktcas', 'kastc', 'kasct', 'katsc', 'katcs',
'kacst', 'kacts', 'kcsta', 'kcsat', 'kctsa', 'kctas', 'kcast',
'kcats']
If you find yourself troubled by duplicates, try fitting your data into a structure with no duplicates like a set:
>>> perms = [''.join(p) for p in permutations('stacks')]
>>> len(perms)
720
>>> len(set(perms))
360
Thanks to #pst for pointing out that this is not what we'd traditionally think of as a type cast, but more of a call to the set() constructor.

You can get all N! permutations without much code
def permutations(string, step = 0):
# if we've gotten to the end, print the permutation
if step == len(string):
print "".join(string)
# everything to the right of step has not been swapped yet
for i in range(step, len(string)):
# copy the string (store as array)
string_copy = [character for character in string]
# swap the current index with the step
string_copy[step], string_copy[i] = string_copy[i], string_copy[step]
# recurse on the portion of the string that has not been swapped yet (now it's index will begin with step + 1)
permutations(string_copy, step + 1)

Here is another way of doing the permutation of string with minimal code based on bactracking.
We basically create a loop and then we keep swapping two characters at a time,
Inside the loop we'll have the recursion. Notice,we only print when indexers reaches the length of our string.
Example:
ABC
i for our starting point and our recursion param
j for our loop
here is a visual help how it works from left to right top to bottom (is the order of permutation)
the code :
def permute(data, i, length):
if i==length:
print(''.join(data) )
else:
for j in range(i,length):
#swap
data[i], data[j] = data[j], data[i]
permute(data, i+1, length)
data[i], data[j] = data[j], data[i]
string = "ABC"
n = len(string)
data = list(string)
permute(data, 0, n)

Stack Overflow users have already posted some strong solutions but I wanted to show yet another solution. This one I find to be more intuitive
The idea is that for a given string: we can recurse by the algorithm (pseudo-code):
permutations = char + permutations(string - char) for char in string
I hope it helps someone!
def permutations(string):
"""
Create all permutations of a string with non-repeating characters
"""
permutation_list = []
if len(string) == 1:
return [string]
else:
for char in string:
[permutation_list.append(char + a) for a in permutations(string.replace(char, "", 1))]
return permutation_list

Here's a simple function to return unique permutations:
def permutations(string):
if len(string) == 1:
return string
recursive_perms = []
for c in string:
for perm in permutations(string.replace(c,'',1)):
recursive_perms.append(c+perm)
return set(recursive_perms)

itertools.permutations is good, but it doesn't deal nicely with sequences that contain repeated elements. That's because internally it permutes the sequence indices and is oblivious to the sequence item values.
Sure, it's possible to filter the output of itertools.permutations through a set to eliminate the duplicates, but it still wastes time generating those duplicates, and if there are several repeated elements in the base sequence there will be lots of duplicates. Also, using a collection to hold the results wastes RAM, negating the benefit of using an iterator in the first place.
Fortunately, there are more efficient approaches. The code below uses the algorithm of the 14th century Indian mathematician Narayana Pandita, which can be found in the Wikipedia article on Permutation. This ancient algorithm is still one of the fastest known ways to generate permutations in order, and it is quite robust, in that it properly handles permutations that contain repeated elements.
def lexico_permute_string(s):
''' Generate all permutations in lexicographic order of string `s`
This algorithm, due to Narayana Pandita, is from
https://en.wikipedia.org/wiki/Permutation#Generation_in_lexicographic_order
To produce the next permutation in lexicographic order of sequence `a`
1. Find the largest index j such that a[j] < a[j + 1]. If no such index exists,
the permutation is the last permutation.
2. Find the largest index k greater than j such that a[j] < a[k].
3. Swap the value of a[j] with that of a[k].
4. Reverse the sequence from a[j + 1] up to and including the final element a[n].
'''
a = sorted(s)
n = len(a) - 1
while True:
yield ''.join(a)
#1. Find the largest index j such that a[j] < a[j + 1]
for j in range(n-1, -1, -1):
if a[j] < a[j + 1]:
break
else:
return
#2. Find the largest index k greater than j such that a[j] < a[k]
v = a[j]
for k in range(n, j, -1):
if v < a[k]:
break
#3. Swap the value of a[j] with that of a[k].
a[j], a[k] = a[k], a[j]
#4. Reverse the tail of the sequence
a[j+1:] = a[j+1:][::-1]
for s in lexico_permute_string('data'):
print(s)
output
aadt
aatd
adat
adta
atad
atda
daat
data
dtaa
taad
tada
tdaa
Of course, if you want to collect the yielded strings into a list you can do
list(lexico_permute_string('data'))
or in recent Python versions:
[*lexico_permute_string('data')]

Here is another approach different from what #Adriano and #illerucis posted. This has a better runtime, you can check that yourself by measuring the time:
def removeCharFromStr(str, index):
endIndex = index if index == len(str) else index + 1
return str[:index] + str[endIndex:]
# 'ab' -> a + 'b', b + 'a'
# 'abc' -> a + bc, b + ac, c + ab
# a + cb, b + ca, c + ba
def perm(str):
if len(str) <= 1:
return {str}
permSet = set()
for i, c in enumerate(str):
newStr = removeCharFromStr(str, i)
retSet = perm(newStr)
for elem in retSet:
permSet.add(c + elem)
return permSet
For an arbitrary string "dadffddxcf" it took 1.1336 sec for the permutation library, 9.125 sec for this implementation and 16.357 secs for #Adriano's and #illerucis' version. Of course you can still optimize it.

Here's a slightly improved version of illerucis's code for returning a list of all permutations of a string s with distinct characters (not necessarily in lexicographic sort order), without using itertools:
def get_perms(s, i=0):
"""
Returns a list of all (len(s) - i)! permutations t of s where t[:i] = s[:i].
"""
# To avoid memory allocations for intermediate strings, use a list of chars.
if isinstance(s, str):
s = list(s)
# Base Case: 0! = 1! = 1.
# Store the only permutation as an immutable string, not a mutable list.
if i >= len(s) - 1:
return ["".join(s)]
# Inductive Step: (len(s) - i)! = (len(s) - i) * (len(s) - i - 1)!
# Swap in each suffix character to be at the beginning of the suffix.
perms = get_perms(s, i + 1)
for j in range(i + 1, len(s)):
s[i], s[j] = s[j], s[i]
perms.extend(get_perms(s, i + 1))
s[i], s[j] = s[j], s[i]
return perms

See itertools.combinations or itertools.permutations.

why do you not simple do:
from itertools import permutations
perms = [''.join(p) for p in permutations(['s','t','a','c','k'])]
print perms
print len(perms)
print len(set(perms))
you get no duplicate as you can see :
['stack', 'stakc', 'stcak', 'stcka', 'stkac', 'stkca', 'satck', 'satkc',
'sactk', 'sackt', 'saktc', 'sakct', 'sctak', 'sctka', 'scatk', 'scakt', 'sckta',
'sckat', 'sktac', 'sktca', 'skatc', 'skact', 'skcta', 'skcat', 'tsack',
'tsakc', 'tscak', 'tscka', 'tskac', 'tskca', 'tasck', 'taskc', 'tacsk', 'tacks',
'taksc', 'takcs', 'tcsak', 'tcska', 'tcask', 'tcaks', 'tcksa', 'tckas', 'tksac',
'tksca', 'tkasc', 'tkacs', 'tkcsa', 'tkcas', 'astck', 'astkc', 'asctk', 'asckt',
'asktc', 'askct', 'atsck', 'atskc', 'atcsk', 'atcks', 'atksc', 'atkcs', 'acstk',
'acskt', 'actsk', 'actks', 'ackst', 'ackts', 'akstc', 'aksct', 'aktsc', 'aktcs',
'akcst', 'akcts', 'cstak', 'cstka', 'csatk', 'csakt', 'cskta', 'cskat', 'ctsak',
'ctska', 'ctask', 'ctaks', 'ctksa', 'ctkas', 'castk', 'caskt', 'catsk', 'catks',
'cakst', 'cakts', 'cksta', 'cksat', 'cktsa', 'cktas', 'ckast', 'ckats', 'kstac',
'kstca', 'ksatc', 'ksact', 'kscta', 'kscat', 'ktsac', 'ktsca', 'ktasc', 'ktacs',
'ktcsa', 'ktcas', 'kastc', 'kasct', 'katsc', 'katcs', 'kacst', 'kacts', 'kcsta',
'kcsat', 'kctsa', 'kctas', 'kcast', 'kcats']
120
120
[Finished in 0.3s]

def permute(seq):
if not seq:
yield seq
else:
for i in range(len(seq)):
rest = seq[:i]+seq[i+1:]
for x in permute(rest):
yield seq[i:i+1]+x
print(list(permute('stack')))

All Possible Word with stack
from itertools import permutations
for i in permutations('stack'):
print(''.join(i))
permutations(iterable, r=None)
Return successive r length permutations of elements in the iterable.
If r is not specified or is None, then r defaults to the length of the iterable and all possible full-length permutations are generated.
Permutations are emitted in lexicographic sort order. So, if the input iterable is sorted, the permutation tuples will be produced in sorted order.
Elements are treated as unique based on their position, not on their value. So if the input elements are unique, there will be no repeat values in each permutation.

This is a recursive solution with n! which accepts duplicate elements in the string
import math
def getFactors(root,num):
sol = []
# return condition
if len(num) == 1:
return [root+num]
# looping in next iteration
for i in range(len(num)):
# Creating a substring with all remaining char but the taken in this iteration
if i > 0:
rem = num[:i]+num[i+1:]
else:
rem = num[i+1:]
# Concatenating existing solutions with the solution of this iteration
sol = sol + getFactors(root + num[i], rem)
return sol
I validated the solution taking into account two elements, the number of combinations is n! and the result can not contain duplicates. So:
inpt = "1234"
results = getFactors("",inpt)
if len(results) == math.factorial(len(inpt)) | len(results) != len(set(results)):
print("Wrong approach")
else:
print("Correct Approach")

With recursive approach.
def permute(word):
if len(word) == 1:
return [word]
permutations = permute(word[1:])
character = word[0]
result = []
for p in permutations:
for i in range(len(p)+1):
result.append(p[:i] + character + p[i:])
return result
running code.
>>> permute('abc')
['abc', 'bac', 'bca', 'acb', 'cab', 'cba']

Yet another initiative and recursive solution. The idea is to select a letter as a pivot and then create a word.
def find_premutations(alphabet):
words = []
word =''
def premute(new_word, alphabet):
if not alphabet:
words.append(word)
else:
for i in range(len(alphabet)):
premute(new_word=word + alphabet[i], alphabet=alphabet[0:i] + alphabet[i+1:])
premute(word, alphabet)
return words
# let us try it with 'abc'
a = 'abc'
find_premutations(a)
Output:
abc
acb
bac
bca
cab
cba

Here's a really simple generator version:
def find_all_permutations(s, curr=[]):
if len(s) == 0:
yield curr
else:
for i, c in enumerate(s):
for combo in find_all_permutations(s[:i]+s[i+1:], curr + [c]):
yield "".join(combo)
I think it's not so bad!

def f(s):
if len(s) == 2:
X = [s, (s[1] + s[0])]
return X
else:
list1 = []
for i in range(0, len(s)):
Y = f(s[0:i] + s[i+1: len(s)])
for j in Y:
list1.append(s[i] + j)
return list1
s = raw_input()
z = f(s)
print z

Here's a simple and straightforward recursive implementation;
def stringPermutations(s):
if len(s) < 2:
yield s
return
for pos in range(0, len(s)):
char = s[pos]
permForRemaining = list(stringPermutations(s[0:pos] + s[pos+1:]))
for perm in permForRemaining:
yield char + perm

from itertools import permutations
perms = [''.join(p) for p in permutations('ABC')]
perms = [''.join(p) for p in permutations('stack')]

def perm(string):
res=[]
for j in range(0,len(string)):
if(len(string)>1):
for i in perm(string[1:]):
res.append(string[0]+i)
else:
return [string];
string=string[1:]+string[0];
return res;
l=set(perm("abcde"))
This is one way to generate permutations with recursion, you can understand the code easily by taking strings 'a','ab' & 'abc' as input.
You get all N! permutations with this, without duplicates.

Everyone loves the smell of their own code. Just sharing the one I find the simplest:
def get_permutations(word):
if len(word) == 1:
yield word
for i, letter in enumerate(word):
for perm in get_permutations(word[:i] + word[i+1:]):
yield letter + perm

This program does not eliminate the duplicates, but I think it is one of the most efficient approaches:
s=raw_input("Enter a string: ")
print "Permutations :\n",s
size=len(s)
lis=list(range(0,size))
while(True):
k=-1
while(k>-size and lis[k-1]>lis[k]):
k-=1
if k>-size:
p=sorted(lis[k-1:])
e=p[p.index(lis[k-1])+1]
lis.insert(k-1,'A')
lis.remove(e)
lis[lis.index('A')]=e
lis[k:]=sorted(lis[k:])
list2=[]
for k in lis:
list2.append(s[k])
print "".join(list2)
else:
break

With Recursion
# swap ith and jth character of string
def swap(s, i, j):
q = list(s)
q[i], q[j] = q[j], q[i]
return ''.join(q)
# recursive function
def _permute(p, s, permutes):
if p >= len(s) - 1:
permutes.append(s)
return
for i in range(p, len(s)):
_permute(p + 1, swap(s, p, i), permutes)
# helper function
def permute(s):
permutes = []
_permute(0, s, permutes)
return permutes
# TEST IT
s = "1234"
all_permute = permute(s)
print(all_permute)
With Iterative approach (Using Stack)
# swap ith and jth character of string
def swap(s, i, j):
q = list(s)
q[i], q[j] = q[j], q[i]
return ''.join(q)
# iterative function
def permute_using_stack(s):
stk = [(0, s)]
permutes = []
while len(stk) > 0:
p, s = stk.pop(0)
if p >= len(s) - 1:
permutes.append(s)
continue
for i in range(p, len(s)):
stk.append((p + 1, swap(s, p, i)))
return permutes
# TEST IT
s = "1234"
all_permute = permute_using_stack(s)
print(all_permute)
With Lexicographically sorted
# swap ith and jth character of string
def swap(s, i, j):
q = list(s)
q[i], q[j] = q[j], q[i]
return ''.join(q)
# finds next lexicographic string if exist otherwise returns -1
def next_lexicographical(s):
for i in range(len(s) - 2, -1, -1):
if s[i] < s[i + 1]:
m = s[i + 1]
swap_pos = i + 1
for j in range(i + 1, len(s)):
if m > s[j] > s[i]:
m = s[j]
swap_pos = j
if swap_pos != -1:
s = swap(s, i, swap_pos)
s = s[:i + 1] + ''.join(sorted(s[i + 1:]))
return s
return -1
# helper function
def permute_lexicographically(s):
s = ''.join(sorted(s))
permutes = []
while True:
permutes.append(s)
s = next_lexicographical(s)
if s == -1:
break
return permutes
# TEST IT
s = "1234"
all_permute = permute_lexicographically(s)
print(all_permute)

This code makes sense to me. The logic is to loop through all characters, extract the ith character, perform the permutation on the other elements and append the ith character at the beginning.
If i'm asked to get all permutations manually for string ABC. I would start by checking all combinations of element A:
A AB
A BC
Then all combinations of element B:
B AC
B CA
Then all combinations of element C:
C AB
C BA
def permute(s: str):
n = len(s)
if n == 1: return [s]
if n == 2:
return [s[0]+s[1], s[1]+s[0]]
permutations = []
for i in range(0, n):
current = s[i]
others = s[:i] + s[i+1:]
otherPermutations = permute(others)
for op in otherPermutations:
permutations.append(current + op)
return permutations

Simpler solution using permutations.
from itertools import permutations
def stringPermutate(s1):
length=len(s1)
if length < 2:
return s1
perm = [''.join(p) for p in permutations(s1)]
return set(perm)

def permute_all_chars(list, begin, end):
if (begin == end):
print(list)
return
for current_position in range(begin, end + 1):
list[begin], list[current_position] = list[current_position], list[begin]
permute_all_chars(list, begin + 1, end)
list[begin], list[current_position] = list[current_position], list[begin]
given_str = 'ABC'
list = []
for char in given_str:
list.append(char)
permute_all_chars(list, 0, len(list) -1)

The itertools module in the standard library has a function for this which is simply called permutations.
import itertools
def minion_game(s):
vow ="aeiou"
lsword=[]
ta=[]
for a in range(1,len(s)+1):
t=list(itertools.permutations(s,a))
lsword.append(t)
for i in range(0,len(lsword)):
for xa in lsword[i]:
if vow.startswith(xa):
ta.append("".join(xa))
print(ta)
minion_game("banana")