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permutation of a string in python

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SHAM3R answered on November 9, 2021 Popularity 10/10 Helpfulness 2/10

answer permutation of a string in python

related Using Python Permutations function on a String

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permutation of a string in python

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-1

# 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)

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Source: www.geeksforgeeks.org

Tags: permutation python string

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Contributed on Dec 30 2021

SHAM3R

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Using Python Permutations function on a String

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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))

Popularity 10/10 Helpfulness 5/10 Language python

Source: softhunt.net

Tags: python py

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Contributed on Apr 10 2022

Outrageous Ostrich

0 Answers Avg Quality 2/10

permutation in a string

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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

Popularity 10/10 Helpfulness 1/10 Language whatever

Source: leetcode.com

Tags: permutation perm

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Contributed on Nov 09 2021

Upset Unicorn

0 Answers Avg Quality 2/10

permutation of a string in python

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