Given target string and argument substring, count how many substrings can be constructed using string characters, repetitions not allowed.
Input : test_str = "geksefokesgergeeks", arg_str = "geeks"
Output : 3
Explanation : "geeks" can be created 3 times using string characters.
Input : test_str = "gefroksefokesgergeeks", arg_str = "for"
Output : 2
Explanation : "for" can be created 2 times using string characters.
Method #1 : Using count() + min() + set()
The combination of above functions can be used to solve this problem. In this, we divide the count of each element of arg string with count of target string character, and using min(), the lowest element is returned. The logic behind is that any element above min would mean the minimum element would miss from that string onwards.
Python3# Python3 code to demonstrate working of# Possible Substring count from String# Using min() + list comprehension + count()# initializing stringtest_str="gekseforgeeks"# printing original stringprint("The original string is : "+str(test_str))# initializing arg stringarg_str="geeks"# using min and count to get minimum possible# occurrence of characterres=min(test_str.count(char)//arg_str.count(char)forcharinset(arg_str))# printing resultprint("Possible substrings count : "+str(res))OutputThe original string is : gekseforgeeksPossible substrings count : 2
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #2 : Using Counter() + list comprehension
This is yet another way in which this task can be performed. In this, we perform the task of counting using Counter() and list comprehension is used to bind the result together, using min() performing similar task as previous method.
Python3# Python3 code to demonstrate working of# Possible Substring count from String# Using Counter() + list comprehensionfromcollectionsimportCounter# initializing stringtest_str="gekseforgeeks"# printing original stringprint("The original string is : "+str(test_str))# initializing arg stringarg_str="geeks"# using Counter to get character frequenciestemp1=Counter(test_str)temp2=Counter(arg_str)res=min(temp1[char]//temp2[char]forcharintemp2.keys())# printing resultprint("Possible substrings count : "+str(res))OutputThe original string is : gekseforgeeksPossible substrings count : 2
The Time and Space Complexity for all the methods are the same:
Time Complexity:O(n)
Auxiliary Space:O(n)
Method 3: using operator.countOf() method
Python3# Python3 code to demonstrate working of# Possible Substring count from Stringimportoperatorasop# initializing stringtest_str="gekseforgeeks"# printing original stringprint("The original string is : "+str(test_str))# initializing arg stringarg_str="geeks"# using min and count to get minimum possible# occurrence of characterres=min(op.countOf(test_str,char)//op.countOf(arg_str,char)forcharinset(arg_str))# printing resultprint("Possible substrings count : "+str(res))OutputThe original string is : gekseforgeeksPossible substrings count : 2
Time Complexity: O(N)
Auxiliary Space : O(1)
Method #4: Using for loop, max_count
Approach:
- Initialize the test_str and arg_str, max count to cnt variable.
- Iterate through the each character in arg_str
- find the count of the character
- compare it with the max count, if it less than the max count
- then assign it to cnt.
- Finally print the count to see the possible substring count.
Python3# Python3 code to demonstrate working of# Possible Substring count from String# Using for loop# Initializing stringtest_str="geksefokesgergeeks"# printing original stringprint("The original string is: ",test_str)# Initializing arg stringarg_str="geeks"# for unique characters in stringarg_str=set(arg_str)# Initializing max countcnt=10000# using for loop to find the count of each character in# arg string and at end combination value will be in cntforiinarg_str:m_cnt=test_str.count(i)ifm_cnt<cnt:cnt=m_cnt# printing resultprint("Possible substrings count: ",cnt)#This code developed by mohan balaji battuOutputThe original string is: geksefokesgergeeksPossible substrings count: 3
Time Complexity: O(N)
Auxiliary Space : O(1)