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Pawn Brotherhood solution in Clear category for Pawn Brotherhood by graffme
def safe_pawns(pawns):
#pionek jest bezpieczny jeśli ma za sobą po skosie innego pionka
#ustawić tablicę z parami pionków
#sprawdzić czy dla danego pionka istnieje pionek zabezpieczający
pawns_indexes = []
for p in pawns:
col = ord(p[0]) - 97
row = int(p[1]) - 1
pawns_indexes.append((col, row))
safe_pawns = 0
for pawn in pawns_indexes:
pawn_col = pawn[0]
pawn_row = pawn[1]
if ((pawn_col -1, pawn_row-1) in pawns_indexes) or ((pawn_col +1, pawn_row-1) in pawns_indexes):
safe_pawns = safe_pawns + 1
return safe_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
Jan. 11, 2017
