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Python | Find all possible substrings after removing k characters

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Examples :

  Input:  geeks, k = 1  Output:  {’gees’,’ eeks’ , ’geks’,’ geek’}  Input:  dog, k = 1  Output:  {’do’,’ dg’, ’og’} 

Approach # 1: Naive Approach
This is a recursive naive approach to find all possible substrings after k characters have been removed. First, we initialize the start, end and index variable with 0, string length and 0 respectively. We create a temporary list, say " temp ", which stores all the output one by one. We start at the first index in temp [] , commit items one by one at that index, and repeat for the rest of the indexes.

# Python3 program for finding all combinations
Number of lines after removing k characters

list = []

 

def findCombinations ( str , temp, start, end, index, k): 

 

if (index = = k): 

  item = ’ ’

 

for j in range (k ): 

item + = temp [j]

  list . append (item)

  return

 

i = start; 

while (i " = end and end - i + 1 " = k - index): 

temp [index] = str [i]

findCombinations ( str , temp, i + 1

  end, index + 1 , k); 

i + = 1

 
Driver code

str = ’geeks’

k = 1

temp = [ 0 ] * ( len ( str ) - k)

s, e = 0 , len ( str ) - 1

  

findCombinations ( str , temp, s, e, 0 , len ( str ) - k)

print ( set ( list ))

Exit:

 {’eeks’,’ gees’, ’geks’,’ geek’} 

Approach # 2 Using Itertools
The Python Itertools module provides a combination () function that takes a string and a length to get all possible string combinations.

# Python3 program to find all combinations
Number of lines after removing k characters

from itertools import combinations

 

def findCombinations ( str , k):

  l = len ( str )

return set ([’’ .join (i) for i in combinations ( str , l - k)])

 
Driver code

str = ’geeks’

k = 1

print (findCombinations ( str , k))

Exit :

 {’geek’,’ eeks’, ’geks’,’ gees’} 

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