Given a string str of length N and a substring pattern of length M, the task is to find the frequency of occurrences of pattern as a substring in the given string. If pattern is present in the string str, then print “Yes” with the count of its occurrence. Otherwise, print “No”.

Examples:

Input: str = “geeksforgeeks”, pattern = “geeks”
Output: 2
Explanation:
The occurrence of the string “geeks” in the string “geeksforgeeks” is at index 0 and 8.
Therefore, the count is 2.

Input: str = “dhimanman”, pattern = “max”
Output: 0

Naive Approach: Refer to the previous post for the simplest approach to solve the problem.
Time Complexity: O(N*M)
Auxiliary Space: O(1)

Approach using KMP Algorithm: Refer to the previous post of this article to solve the problem using KMP algorithm.
Time Complexity: O(N + M)
Auxiliary Space: O(M)

Approach using Regular Expression: Follow the steps below to solve the problem:

• Form the regular expression of the string pattern using regex() function.
• Create a smatch M using function smatch().
• Check the presence of the string pattern in the string str using function regex_match() as:

regex_search(str, m, c)
where,
str is the given string,
m is smatch,
c is the regular expression of the string pattern.

• In the above steps, if the function regex_match() returns True, then print “Yes” and find the occurrence of the string pattern. Otherwise, print “No”.
• Create a variable numberOfMatches of data type ptrdiff_t to store the count of occurrence.
• Find the numberOfMatches using function regex_iterator() function as:

ptrdiff_t numberOfMatches = std::distance(sregex_iterator(S.begin(), S.end(), c), sregex_iterator())

• Print the count of occurrence in the above steps as the result.

Below is the implementation of the above approach:

Yes
Frequency of string geeks is 2

Time Complexity: O(N + M)
Auxiliary Space: O(M)

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