def uniquePaths(self, m: int, n: int) -> int:
"""
Consider each move as 1-bit string, say for m = 3 and n = 2,
possible path could be RRD, RDR, DRR. There should be at least one D and two R.
So, we can find all combinations where there is one D in three moves.
Below is an implementation of closed formula for n chooses k = n!/(n-k)!k!.
"""
total_moves = m - 1 + n - 1
down_moves = n - 1
return mf(total_moves) // (mf(total_moves - down_moves) * mf(down_moves))
def decimal_to_any(num, base):
fractional = False
if isinstance(num, float):
points = 5
num = [int(i) for i in str(num).split(".")]
integer = num[0]
length = len(str(num[1]))
fractional = float(str("0.") + str(num[1]))
else:
integer = num
ans = ""
# Decimal part conversion
while integer != 0:
intermediate = integer % base
intermediate = ref[intermediate]
ans = str(intermediate) + ans
integer //= base
# Floating part conversion
if fractional:
ans += "."
while points:
fractional, i = mf(fractional * base)
fractional = round(fractional, length)
i = ref[int(i)]
ans += i
points -= 1
return ans
def binomail(n, r):
return (mf(n) / (mf(r) * mf(n - r)))