Sometimes, while working with Python, we have a problem in which we compute how many characters are present in string. But sometimes, we can have a problem in which we need to get all characters that occur in atleast K Strings. Let’s discuss certain ways in which this task can be performed.
Method #1: Using set() + Counter() + items() + loop The combination of above functions can be used to perform this particular task. In this, we find the count using Counter and set is used to limit the character duplicacy.
Python3
# Python3 code to demonstrate # Characters which Occur in More than K Strings# using set() + Counter() + loop + items()from collections import Counterfrom itertools import chaindef char_ex(strs, k): temp = (set(sub) for sub in strs) counts = Counter(chain.from_iterable(temp)) return {chr for chr, count in counts.items() if count >= k}# Initializing listtest_list = ['Gfg', 'ise', 'for', 'Geeks']# printing original listprint("The original list is : " + str(test_list))# Initializing KK = 2# Characters which Occur in More than K Strings# using set() + Counter() + loop + items()res = char_ex(test_list, K) # printing result print ("Filtered Characters are : " + str(res)) |
The original list is : ['Gfg', 'ise', 'for', 'Geeks']
Filtered Characters are : {'e', 'G', 's', 'f'}
Time complexity: O(nm) where n is the number of strings in the list and m is the average length of the strings.
Auxiliary Space complexity: O(nm) as well, as it uses a set and a Counter object to store characters and their counts.
Method #2: Using dictionary comprehension + set() + Counter() The combination of these functions perform this task in similar way as above. The difference is just that we use dictionary comprehension to give a compact solution to problem.
Python3
# Python3 code to demonstrate # Characters which Occur in More than K Strings# using set() + Counter() + dictionary comprehensionfrom collections import Counter# Initializing listtest_list = ['Gfg', 'ise', 'for', 'Geeks']# printing original listprint("The original list is : " + str(test_list))# Initializing KK = 2# Characters which Occur in More than K Strings# using set() + Counter() + dictionary comprehensionres = {key for key, val in Counter([ele for sub in test_list for ele in set(sub)]).items() if val >= K} # printing result print ("Filtered Characters are : " + str(res)) |
The original list is : ['Gfg', 'ise', 'for', 'Geeks']
Filtered Characters are : {'e', 'G', 's', 'f'}
Time complexity: O(n), where n is the total number of characters in the test_list.
Auxiliary Space: O(n), as the Counter object stores the count of all characters in the test_list.
Method #3: Using nested for loops
Step-by-step approach:
- Join all the strings of test_list and store in x
- Remove duplicate characters from x and store again in x using list(),set() methods
- Count the occurrence each character of x in each string of test_list, and check if the count is greater than or equal to K
- If yes append to output list res
- Display res
Python3
# Python3 code to demonstrate# Characters which Occur in More than K Strings# Initializing listtest_list = ['Gfg', 'ise', 'for', 'Geeks']# printing original listprint("The original list is : " + str(test_list))# Initializing KK = 2res=[]# Characters which Occur in More than K Stringsx="".join(test_list)x=list(set(x))for i in x: c=0 for j in test_list: if i in j: c+=1 if(c>=K): res.append(i)# printing resultprint ("Filtered Characters are : " + str(res)) |
The original list is : ['Gfg', 'ise', 'for', 'Geeks'] Filtered Characters are : ['G', 'f', 's', 'e']
Time Complexity: O(M*N) ,where M is the length of test_list N – length of each string in test_list
Auxiliary Space: O(N) ,where N is the length of res
