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Sometimes, while working with Python strings, we can have a problem in which we need to check for maximum occurring substring from strings list. This can have application in DNA sequencing in Biology and other application. Let's discuss certain way in which this task can be performed.
Method 1 : Using regex() + groupby() + max() + lambda
The combination of above functionalities can be used to solve this particular problem. In this, we first extract the sequences using regex function. Then the counter grouping is performed using groupby(). The last step is extracting maximum which is done using max() along with lambda function.
The original string is : gfghsisbjknlmkesbestgfgsdcngfgcsdjnisdjnlbestdjsklgfgcdsbestbnjdsgfgdbhisbhsbestdkgfgb The original list is : ['gfg', 'is', 'best'] Maximum frequency substring : gfg
Time complexity: O(n), where n is the length of the input string. The time complexity of regex(), groupby(), and max() is O(n).
Auxiliary space: O(k), where k is the length of the input list. This is the space needed to store the list of substrings. The space complexity of regex(), groupby(), and max() is O(1).
Method 2: Using count() and max() methods
count() returns the occurrence of a particular element in a sequence and the max() method returns the maximum of that.
The original string is : gfghsisbjknlmkesbestgfgsdcngfgcsdjnisdjnlbestdjsklgfgcdsbestbnjdsgfgdbhisbhsbestdkgfgb The original list is : ['gfg', 'is', 'best'] Maximum frequency substring : gfg
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 3: Using re.findall() + Counter
This is an alternate approach that uses re.findall() and Counter module. In this, we extract the sequence using re.findall() and count the occurrence of each element using Counter() from collections module.
The original string is : gfghsisbjknlmkesbestgfgsdcngfgcsdjnisdjnlbestdjsklgfgcdsbestbnjdsgfgdbhisbhsbestdkgfgb The original list is : ['gfg', 'is', 'best'] Maximum frequency substring : gfg
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 4 : Using operator.countOf() and max() methods
The original string is : gfghsisbjknlmkesbestgfgsdcngfgcsdjnisdjnlbestdjsklgfgcdsbestbnjdsgfgdbhisbhsbestdkgfgb The original list is : ['gfg', 'is', 'best'] Maximum frequency substring : gfg
Time Complexity : O(n)
Auxiliary Space : O(n)
Method 5: Using a dictionary to count occurrences
In this approach, we can use a dictionary to count the occurrences of each substring in the list. We can iterate over the string and for each substring in the list, we can count the number of occurrences of that substring in the string and update the count in the dictionary. Finally, we can find the substring with the maximum count in the dictionary.
Approach:
Example:
The original string is : gfghsisbjknlmkesbestgfgsdcngfgcsdjnisdjnlbestdjsklgfgcdsbestbnjdsgfgdbhisbhsbestdkgfgb The original list is : ['gfg', 'is', 'best'] Maximum frequency substring : gfg
Time complexity: O(n*m), where n is the length of the string and m is the total number of substrings in the list.
Auxiliary space: O(m), where m is the total number of substrings in the list.
Method 6: Using itertools.product() and count()
Example:
The original string is : gfghsisbjknlmkesbestgfgsdcngfgcsdjnisdjnlbestdjsklgfgcdsbestbnjdsgfgdbhisbhsbestdkgfgb The original list is : ['gfg', 'is', 'best'] Maximum frequency substring : gfg
Time complexity: O(n*m^2), where n is the length of test_list and m is the maximum length of a substring in test_list.
Auxiliary space: O(m)