First clear solution for Pawn Brotherhood by welv - python coding challenges

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First solution in Clear category for Pawn Brotherhood by welv

import string


def isPawnSafe(pawns, pawn):
    l = list(string.ascii_lowercase)
    row = int(pawn[1]) - 1
    leftLetter = l[l.index(pawn[0]) - 1]
    rightLetter = l[l.index(pawn[0]) + 1]

    return (leftLetter + str(row)) in pawns or (rightLetter + str(row)) in pawns



def safe_pawns(pawns):
    safePawnsCount = 0
    for p in pawns:
        if(isPawnSafe(pawns, p)):
            safePawnsCount += 1
    return safePawnsCount



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

Jan. 20, 2017

Look up in the Dict()

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