Question

C++ LANGUAGE

/**

* Given an input string, count the number of words ending

* in 'y' or 'z' -so the 'y' in "heavy" and the 'z' in "fez"

* count, but not the 'y' in "yellow". Make sure that your

* comparison is not case sensitive. We'll say that a y

* or z is at the end of a word if there is not an alphabetic

* letter immediately following it.

*

* Do not use any string functions except for substr(),

* at(), and size(). You may use isalpha from cctype as well.

*

*  

* Here are some other examples:

* - "fez day" -> 2

* - "day fez" -> 2

* - "day fyyyz" -> 2  

*/

int endzy(const string& str)

{

int result;

  

// Your code goes here

  

return result;

}

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Answer 1

/**
* Given an input string, count the number of words ending
* in 'y' or 'z' -so the 'y' in "heavy" and the 'z' in "fez"
* count, but not the 'y' in "yellow". Make sure that your
* comparison is not case sensitive. We'll say that a y
* or z is at the end of a word if there is not an alphabetic
* letter immediately following it.
*
* Do not use any string functions except for substr(),
* at(), and size(). You may use isalpha from cctype as well.
*
*
* Here are some other examples:
* - "fez day" -> 2
* - "day fez" -> 2
* - "day fyyyz" -> 2
*/
int endzy(const string &str) {
    int result = 0;
    char ch;
    for (int i = 0; i < str.size(); ++i) {
        ch = str[i];
        if (ch >= 'A' && ch <= 'Z') {
            ch += 32;
        }
        if (i == str.size() - 1 || str[i + 1] == ' ') {
            if (ch == 'y' || ch == 'z') {
                result++;
            }
        }
    }
    return result;
}