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First solution in Clear category for Pawn Brotherhood by GiMiX311
def safe_pawns(pawns: set) -> int:
# This commented bit of code is what that convoluted bit of code that returns the result equivalates.
# count = 0 # set up counter
# for x in pawns: # iterate over each coordinate
# # let a and b take possible coordinates of pieces that can guard the current piece
# a, b = chr(ord(x[0])-1) + str(int(x[1])-1), chr(ord(x[0])+1) + str(int(x[1])-1)
# if a in pawns or b in pawns: # check if such pieces exist
# count +=1 # increment counter if they exist
return sum(any(x in pawns for x in (chr(ord(i[0])-1) + str(int(i[1])-1), chr(ord(i[0])+1) + str(int(i[1])-1))) for i 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!")
April 25, 2020
