How to use break_it method in avocado

Best Python code snippet using avocado_python

Day10_exercises.py

Source:Day10_exercises.py

...53#  add a new method break_it with two default parameters max_lines=-1 and drop equal to "yeah", which is similar to sing, but adds a drop after each word .54# Example: 55# zrap = Rap("ZiemeÄ¼meita", "Jumprava", ["GÄju meklÄt ziemeÄ¼meitu","56# Garu, tÄlu ceÄ¼u veicu"])57# zrap.break_it(1, "yah")58# ZiemeÄ¼meita - Jumprava59# GÄju YAH meklÄt YAH ziemeÄ¼meitu YAH60class Rap(Song):61    def break_it(self, max_lines=-1, drop="yeah"):62        print(f"{self.title} - {self.author}")63        if max_lines == -1:64            max_lines = len(self.lyrics)65        for line in self.lyrics[:max_lines]:66            words = line.split()67            new_line = f" {drop} ".join(words) + " " + drop68            print(new_line)69        return self70zrap = Rap("ZiemeÄ¼meita", "Jumprava", ["GÄju meklÄt ziemeÄ¼meitu","Garu, tÄlu ceÄ¼u veicu"])...

break_a_palindrome.py

Source:break_a_palindrome.py

1class Palindrome:2    def break_it(self, pal: str) -> str:3        """4        Conditions:5        -----------6        1. If length of palindrome is one return empty string.7        2. If even length string, we found a char that is not "a" replace it8           with a and return.9        3. If odd len string, find a char that is not "a" and if it is not middle of10           string replace with "a"11        4. If all "a" in string, replace the last char with b.12        Approach: String manipulation13        Time Complexity: O(N)14        Space Complexity: O(1)15        :param pal:16        :return:17        """18        if len(pal) == 1:19            return ""20        p1, p2 = 0, len(pal) - 121        while p1 < p2:22            if pal[p1] != "a":23                pal = pal[:p1] + "a" + pal[p1 + 1:]24                return pal25            p1 += 126            p2 -= 127        return pal[:-1] + "b"28if __name__ == "__main__":29    palindrome = Palindrome()30    print(palindrome.break_it("aaaa"))31    print(palindrome.break_it("aba"))32    print(palindrome.break_it("a"))33    print(palindrome.break_it("bb"))...

Longest Common Subsequence.py

Source:Longest Common Subsequence.py

2    def longestCommonSubsequence(self, text1: str, text2: str) -> int:3            4        explored_strings={}5        6        def break_it(str1, str2):7            8            if (str1, str2) in explored_strings:9                return explored_strings[(str1, str2)]10            11            if len(str1)==0 or len(str2)==0:12                return 013            if str1[0]==str2[0]:14                 explored_strings[(str1, str2)]=1+break_it(str1[1:], str2[1:])15            else:16                 explored_strings[(str1, str2)]=max(break_it(str1[1:], str2), break_it(str1, str2[1:]))17            return explored_strings[(str1, str2)]18        19        return break_it(text1, text2)...

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