One line clear solution for Pawn Brotherhood by mishasha13 - python coding challenges

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One line solution in Clear category for Pawn Brotherhood by mishasha13

def safe_pawns(pawns: set) -> int:
    return sum( ( (chr(ord(p[0])-1)+str(int(p[1])-1)) in pawns or (chr(ord(p[0])+1)+str(int(p[1])-1)) in pawns) for p in pawns)

if __name__ == '__main__':
    #These "asserts" using only for self-checking and not necessary for auto-testing
    assert safe_pawns({"b4", "d4", "f4", "c3", "e3", "g5", "d2"}) == 6
    assert safe_pawns({"b4", "c4", "d4", "e4", "f4", "g4", "e5"}) == 1
    print("Coding complete? Click 'Check' to review your tests and earn cool rewards!")

Aug. 8, 2018

Comments:

Galeon13 on Aug. 9, 2018, 9:31 a.m.
<p>Nice idea!</p>

hyl3rid on Nov. 12, 2018, 12:43 a.m.
<p>List comprehension is being used. The function is placed before the for loop and checks if the row below has pawns which protect the current pawns position number. Columns use chr and ord to convert the letter into a number, reduce the number and convert it back into a letter. </p>

Look up in the Dict()

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