Very easy method clear solution for Long Repeat by ail531 - python coding challenges

Enable Javascript in your browser and then refresh this page, for a much enhanced experience.

Very easy method solution in Clear category for Long Repeat by ail531

def long_repeat(line: str) -> int:
    if line:
        previous_letter = line[0]  ## As at least one letter is 
        count = 1
        max_count = 1

        for letter in line[1:]:
            if letter == previous_letter:        
                count +=1
                if count > max_count:
                    max_count = count
            else:
                count = 1 
            previous_letter = letter
        return max_count
    else:
        return 0

if __name__ == '__main__':
    #These "asserts" using only for self-checking and not necessary for auto-testing
    assert long_repeat('sdsffffse') == 4, "First"
    assert long_repeat('ddvvrwwwrggg') == 3, "Second"
    assert long_repeat('abababaab') == 2, "Third"
    assert long_repeat('') == 0, "Empty"
    print('"Run" is good. How is "Check"?')

Sept. 4, 2019

Comments:

cured on March 10, 2020, 11:59 p.m.
<p>i prefer to write all constraints at first and then the logic in your example it can look like:</p> <pre class='brush: python'>def long_repeat(line: str) -> int: if not line: return 0 previous_letter = line[0] ## As at least one letter is count = 1 max_count = 1 for letter in line[1:]: if letter == previous_letter: count +=1 if count > max_count: max_count = count else: count = 1 previous_letter = letter return max_count </pre> <p>in this case the code is not overloaded with else block</p>

Strings Theory

0 %

You ClassRoom Demo is Ready With the new tool CheckiO ClassRoom, you will be able to analyze your students' progress, have your own leaderboard, change missions order on the map, and so on.