First uncategorized solution for Goes Right After (simplified) by HubiJohn98 - python coding challenges

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First solution in Uncategorized category for Goes Right After (simplified) by HubiJohn98

def goes_after(word: str, first: str, second: str) -> bool:
    if (first not in word) or (second not in word) or (first == second):
        return False
    for letter in word[:-1]:
        if (letter == first) and (word[word.index(letter)+1] == second):
            return True
    return False


print("Example:")
print(goes_after("world", "w", "o"))

# These "asserts" are used for self-checking
assert goes_after("world", "w", "o") == True
assert goes_after("world", "w", "r") == False
assert goes_after("world", "l", "o") == False
assert goes_after("list", "l", "o") == False
assert goes_after("", "l", "o") == False
assert goes_after("list", "l", "l") == False
assert goes_after("world", "d", "w") == False

print("The mission is done! Click 'Check Solution' to earn rewards!")

Dec. 13, 2022

Comments:

126083 on Jan. 29, 2023, 5:37 p.m.
<p>Było można zrobić to w prostszy sposób :)</p> <pre class='brush: python'>def goes_after(word: str, first: str, second: str) -> bool: slowo=first+second if slowo in word: return True else: return False </pre>

Strings and Integers

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