How to solve this problem with binary search?

• I used hash table with powers of divider as keys, their occurrences as values and add them up.
EG.75/3=25 map = {9: 2, 3: 2, 1: 1}
9+9+3+3+1=25
Please tell me how to solve this problem with binary search, not bit manipulation. Many thanks!

``````class Solution:
# @param {integer} dividend
# @param {integer} divisor
# @return {integer}
def divide(self, dividend, divisor):
flag = True
if (dividend>0 and divisor<0) or (dividend<0 and divisor>0):
flag = False
dividend = abs(dividend)
divisor = abs(divisor)

if divisor == dividend:
return 1 if flag else -1
if divisor > dividend:
return 0

if divisor == 1:
if dividend == pow(2,31) and flag:
return dividend-1
return dividend if flag else -dividend
#Construct a map with low_pow: cur_num as key value pair
#EG.75/3=25 map = {9: 2, 3: 2, 1: 1}
#9+9+3+3+1=25
hash_table = {}
n = 0
while pow(divisor,n) < dividend:
n += 1
low_pow = pow(divisor, n-1)
while dividend >= divisor:
cur_num = 0
while dividend >= low_pow and n > 0:
dividend -= low_pow
cur_num += 1
n -= 1
low_pow = pow(divisor, n-1)
if cur_num != 0:
hash_table[low_pow] = cur_num
res = 0
for key in hash_table.keys():
while hash_table[key] != 0:
res += key
hash_table[key] -= 1
if flag:
return res
else:
return -res``````

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