PYTHON

Using Python Permutations function on a String

from itertools import permutations 
string="SOFT"
a=permutations(string) 
for i in list(a): 
   # join all the letters of the list to make a string 
   print("".join(i))

python all permutations of a string

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

permutation of a string in python

# Function to find permutations of a given string
from itertools import permutations
  
def allPermutations(str):
       
     # Get all permutations of string 'ABC'
     permList = permutations(str)
  
     # print all permutations
     for perm in list(permList):
         print (''.join(perm))
        
# Driver program
if __name__ == "__main__":
    str = 'ABC'
    allPermutations(str)

Permutation in python

from itertools import permutations
from itertools import combinations
p = permutations([1,2,4]) # or  permutations([1, 2, 3], 2)
for i in p:
    print(i)
c = combinations([1,2,3],2)
for j in c:
    print(j)
    
    

permutation in a string

import collections
class Solution:
    def checkInclusion(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: bool
        """
        '''
        Apprach 3:
        Just changing the collections.Counter to collections.defaultdict
        improved the speed
        94.36%, Nice. :)
        
        '''
        ctr1 = collections.defaultdict(int)
        ctr2 = collections.defaultdict(int)
        for x in s1: ctr1[x] += 1
        for x in s2[:len(s1)]: ctr2[x] += 1
            
        i = 0; j = len(s1)
        
        while j < len(s2):
            if ctr2 == ctr1: return True
            ctr2[s2[i]] -= 1
            if ctr2[s2[i]] < 1: ctr2.pop(s2[i]); 
            ctr2[s2[j]] = ctr2.get(s2[j], 0) + 1
            i += 1; j += 1
            
        return ctr2 == ctr1
        
        '''
        Approach 2:
        Use the same counter but much more effeciently
        63% beated, okay not bad..
        two pointers i(front) j(last), delete and add accordingly and check for the counts
        
        # Code Below
        '''
        ctr1 = collections.Counter(s1)
        ctr2 = collections.Counter(s2[:len(s1)])
            
        i = 0; j = len(s1)
        
        while j < len(s2):
            if ctr2 == ctr1: return True
            ctr2[s2[i]] -= 1
            if ctr2[s2[i]] < 1: ctr2.pop(s2[i]); 
            ctr2[s2[j]] = ctr2.get(s2[j], 0) + 1
            i += 1; j += 1
            
        return ctr2 == ctr1
    
        
        '''
        1. Approach 1
        Things i Missed: 
            - Had the len(s2) - len(s1) + 1: forgot to add the 1 in the if condition 
        Ultra easy solution
        Slice and check for the equality
        5% beat, yuck !! 
        # Code below
       '''
        ctr1 = collections.Counter(s1)
        i = 0
        while i < len(s2) - len(s1) + 1:
            if s2[i] in ctr1:
                ctr2 = collections.Counter(s2[i: i+len(s1)])
                if ctr1 == ctr2: return True
            i += 1
        return False