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chr() ord() solution in Clear category for Pawn Brotherhood by lucas.stonedrake
def safe_pawns(pawns: set) -> int:
result = 0
#for each pawn, derive the coordinates of the two squares which can cover it
for p in pawns:
friend1 = chr(ord(p[0])-1) + chr(ord(p[1])-1) # column left, one row behind
friend2 = chr(ord(p[0])+1) + chr(ord(p[1])-1) # column right, one row behind
if friend1 in pawns or friend2 in pawns: # if either is present
result += 1
return result
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 19, 2021
