/**
 *  @author Huy
 *  @date 09.23.2018
 *  @file h05.cpp
 */
#include <string>
#include <iostream>
using namespace std;

string STUDENT = "hnguyen771"; // Add your Canvas/occ-email ID

#include "h05.h"

// Place your function definitions in this file.
/**
    zipZap(str) removes the middle letters from all "zip" or "zap" strings.

    @param str the input string.
    @return Look for patterns like "zip" and "zap" in the input string:
        any substring of length 3 that starts with "z" and ends with "p".
        Return a string where for all such words, the middle letter
        is gone, so "zipXzap" returns "zpXzp".
        <ul>
        <li>zipZap("zipXzap") returns "zpXzp"
        <li>zipZap("zopzop") returns "zpzp"
        <li>zipZap("zzzopzop") returns "zzzpzp"
        </ul>
*/
string zipZap(const string& str)
{
    string result;
    // 1. Save the length (len) and create an index (i = 0)
    // 2. If len is less than 3, then we just return the input str
    // 3. Write a while loop that goes while i is less than len – 2
    // 4. Inside the loop, extract a substring of 3 (word)
    // 5. If word starts with 'z' and ends with 'p' then:
    //      - add "zp" to the output string
    //      - move forward three characters (to skip the "ip")
    // 6. Otherwise
    //      - add the first character of word to the output string
    //      - move forward one character
    // 7. After the loop, add the remaining characters, to the output

    size_t i;
    size_t len = str.size();
    i = 0;

    if (len < 3)
    {
        return str;
    }

    while (i < len - 2)
    {
        string word = str.substr(i, 3);
        if (word.front() == 'z' && word.back() == 'p')
        {
            result = result + "zp";
            i = i + 3;
        }
        else
        {
            result = result + word.at(i);
            i = i + 1;
        }
    }


    return result;
}
/**
    countCode(str) counts all occurences of the "code" pattern in str.

    @param str the input string.
    @return the number of times that the string "code" appears
        anywhere in the given input string, except that we'll accept
        <b>any</b> letter for the 'd', so "cope" and both "cooe" count.
        <ul>
        <li>countCode("aaacodebbb") returns 1
        <li>countCode("codexxcode") returns 2
        <li>countCode("cozexxcope") returns 2
        </ul>
*/

int countCode(const string& str)
{
    int result = 0;

    // Loop through str, grabbing a 4-character substring each time
    size_t i;
    size_t len = str.size();
    for (i = 4; i <= len; i++)
    {
        string word = str.substr(i - 4, 4); // 4 or fewer characters
        string front = word.substr(0, 2);
        string back = word.substr(3);
        if (front == "co" && back == "e")
        {
            result = result + 1;
        }
    }


    return result;
}

/**
    everyNth(str, n) calculates every nth character.

    @param str the input string to check.
    @param n the n-th character to use
    @return the string made starting with char 0, and then
        every n-th char of the string. So, if n is 3,
        use char 0, 3, 6, and so on. If n is
        less than 1, return the empty string
        <ul>
        <li>everyNth("Miracle", 2) returns "Mrce"
        <li>everyNth("abcdefg", 2) returns "aceg"
        <li>everyNth("abcdefg", 3) returns "adg"
        </ul>
*/
string everyNth(const string& str, int n)
{
    string result;
    size_t i;
    size_t len = str.size();

    for (i = 0; i < len; i = i + n)
    {
        result = result + str.at(i);
    }

    return result;
}
/**
    prefixAgain(str, n) returns true when the prefix(0,n) appears again in the string.

    @param str the input string.
    @param n the number of characters to count for the prefix.
    @return consider the prefix string made of the first n characters of
        the input string. Does that prefix string appear somewhere else
        in the string? Assume that the string is not empty and that n
        is in the range 1..inStr.length().
        <ul>
        <li>prefixAgain("abXYabc", 1) returns true
        <li>prefixAgain("abXYabc", 2) returns true
        <li>prefixAgain("abXYabc", 3) returns false
        </ul>
*/
bool prefixAgain(const string& str, int n)
{
    // 1. Extract the first n characters into prefix
    // 2. Start looping at 1, extracting n characters into word
    //      - if word is equal to prefix, return true
    // 3. If we get through the loop, return false
    return str.substr(1).find(str.substr(0, n)) != string::npos;
}





/////////// STUDENT TESTING //////////////////////
int run()
{
    cout << "Student tests" << endl;
    return 0;
}