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’}
Python | Find all possible substrings after removing k characters find: Questions
Python | Find all possible substrings after removing k characters Python functions: Questions
