Write a python program to print all the permutations of a string in lexicographical order. 

Examples:

Input  : python
Output : hnopty
hnopyt
hnotpy
hnotyp
hnoypt
......
ytpnho
ytpnoh
ytpohn
ytponh

Input  : xyz
Output : xyz
xzy
yxz
yzx
zxy
zyx

Method 1: Using the default library itertools function permutations. permutations function will create all the permutations of a given string and then we sort the result to get our desired output. 

from itertools import permutations

def lexicographical_permutation(str):
    perm = sorted(''.join(chars) for chars in permutations(str))
    for x in perm:
        print(x)
        
str ='abc'
lexicographical_permutation(str)

from itertools import permutations

​

def lexicographical_permutation(str):

    perm = sorted(''.join(chars) for chars in permutations(str))

    for x in perm:

        print(x)

str ='abc'

lexicographical_permutation(str)

Output :

abc
acb
bac
bca
cab
cba

Method 2:

• First we create a loop that will run n! ties where n is the length of the string as there will be n! permutations.
• Every iteration prints the string and finds its next larger lexicographical permutation to be printed in the next iteration.
• The next higher permutation is found as :-
• Let the string is called str, find the smallest index i such that all elements in str[i…end] are in descending order.
• If str[i…end] is the entire sequence, i.e. i == 0, then str is the highest permutation. So we simply reverse the entire string to get the smallest permutation which we consider as the next permutation.
• If i > 0, then we reverse str[i…end].
• Then we look for the smallest element in str[i…end] that is greater than str[i – 1] and swap its position with str[i – 1].
• This is then the next permutation.

# import library
from math import factorial
  
def lexicographical_permutations(str):
     
    # there are going to be n ! permutations where n = len(seq)
    for p in range(factorial(len(str))):         
        print(''.join(str))  
     
        i = len(str) - 1
        
        # find i such that str[i:] is the largest sequence with
        # elements in descending lexicographic order
        while i > 0 and str[i-1] > str[i]:       
            i -= 1
  
        # reverse str[i:]
        str[i:] = reversed(str[i:]) 
         
  
        if i > 0:
             
            q = i
            # find q such that str[q] is the smallest element
            # in str[p:] such that str[q] > str[i - 1]
            while str[i-1] > str[q]:  
                q += 1
            
            # swap str[i - 1] and str[q]
            temp = str[i-1] 
            str[i-1]= str[q]
            str[q]= temp
             
  
s = 'abcd'
s = list(s)
s.sort()
lexicographical_permutations(s)

# import library

from math import factorial

def lexicographical_permutations(str):

    # there are going to be n ! permutations where n = len(seq)

    for p in range(factorial(len(str))):         

        print(''.join(str))  

        i = len(str) - 1

        # find i such that str[i:] is the largest sequence with

        # elements in descending lexicographic order

        while i > 0 and str[i-1] > str[i]:       

            i -= 1

        # reverse str[i:]

        str[i:] = reversed(str[i:]) 

        if i > 0:

            q = i

            # find q such that str[q] is the smallest element

            # in str[p:] such that str[q] > str[i - 1]

            while str[i-1] > str[q]:  

                q += 1

            # swap str[i - 1] and str[q]

            temp = str[i-1] 

            str[i-1]= str[q]

            str[q]= temp

s = 'abcd'

s = list(s)

s.sort()

lexicographical_permutations(s)

Output : 

abcd
abdc
acbd
acdb
adbc
adcb
bacd
badc
bcad
bcda
bdac
bdca
cabd
cadb
cbad
cbda
cdab
cdba
dabc
dacb
dbac
dbca
dcab
dcba

Time Complexity: O(n*n!)
Auxiliary Space: O(1)