Number of happy and sad subsequences in a string with three sections X16298

In this exercise you have to carry out several tasks.

Firstly, you’ll have to implement a function that receives a string s holding some very particular conditions: s is formed using three different characters c₁, c₂, c₃. Moreover, at the beginning of s there are one or more occurrences of character c₁, next there are one or more occurrences of character c₂, and finally there are one or more occurrences of character c₃. The function will also be given three extra integer parameters n₁, n₂, n₃ by reference, where it has to assign the number of occurrences of c₁, c₂, c₃, respectively. This is the header:

// Pre: s is formed with three different characters c1,c2,c3, and is of the form c1...c1c2...c2c3...c3.
// Post: n1, n2, n3 are the number of occurrences of c1, c2, c3 in s, respectively.
void numberOccurrences(const string &s, int &n1, int &n2, int &n3);

Note: the private tests of this exercise are big and designed so that one needs an implementation with logarithmic cost of numberSubsequences. A slow implementation will let you overcome the public tests and get half the score.

Secondly, you’ll have to implement a function that receives a string s holding one of the following conditions:

1. s begins with one or more occurrences of :, followed by one or more occurrences of −, followed by one or more occurrences of ), and there are no more characters in s.
2. s begins with one or more occurrences of (, followed by one or more occurrences of −, followed by one or more occurrences of :, and there are no more characters in s.
3. s begins with one or more occurrences of :, followed by one or more occurrences of −, followed by one or more occurrences of (, and there are no more characters in s.
4. s begins with one or more occurrences of ), followed by one or more occurrences of −, followed by one or more occurrences of :, and there are no more characters in s.

In former cases (1) and (2), the function will return the number of happy subsequences of s, and in former cases (3) and (4), the function will return the number of sad subsequences of s. A happy subsequence is one having three characters, where those three characters are, either :-) or (-:, in the given order. A sad subsequence is one having three characters, where those three characters are, either :-( or )-:, in the given order. This is the header:

// Pre: s begins with one or more occurrences of a character c1, followed by one or more
//      occurrences of a character c2, followed by one or more occurrences of a character c3,
//      and there are no more characters in s.
//      moreover, either c1c2c3 = ":-)" or c1c2c3 = "(-:" or c1c2c3 = ":-(" or c1c2c3 = ")-:".
// Post: If c1c2c3 = ":-)" or c1c2c3 = "(-:", the function returns the number of happy subsequences.
//       If c1c2c3 = ":-(" or c1c2c3 = ")-:", the function returns the number of sad subsequences.
int numberHappyOrSadSubsequences(const string &s);

Former function must conveniently use function numberOccurrences described at the beginning. Otherwise, the delivery will be invalidated.

Observation You only need to submit the required procedure; your main program will be ignored.

Observation

Grading up to 10 points:

• Slow solution: 5 points.
• Fast solution: 10 points.

We understand as a fast solution one which is correct, with logarithmic cost and which passes the public and private tests. We understand as slow solution one which is not fast, but it is correct and passes the public tests.