[Python] Easiest but not efficient solution[Python] Easiest but not efficient solution
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A Billion Bowling Pins
Week 5, 2026
def calc_pin_rows(no_of_pins):
rows = 0
n = 1
if no_of_pins == 1:
return 1
while no_of_pins > 0:
if no_of_pins - n < 0:
break
rows += 1
no_of_pins -= n
n += 1
return rows
if __name__ == "__main__":
no_of_pins = 1000000000 #for billion
rows = calc_pin_rows(no_of_pins)
print(f"Number of rows needed for {no_of_pins} pins: {rows}")
