Hard
Minimum Operations to Equalize Binary String — Python
Full explanation · Time O(n) · Space O(1)
# Time: O(n)
# Space: O(1)
# math
class Solution(object):
def minOperations(self, s, k):
"""
:type s: str
:type k: int
:rtype: int
"""
def ceil_divide(a, b):
return (a+b-1)//b
zero = s.count('0')
if len(s) == k:
return 0 if zero == 0 else 1 if zero == len(s) else -1
result = float("inf")
if (k&1) == (zero&1):
i = max(ceil_divide(zero, k), ceil_divide(len(s)-zero, len(s)-k))
if (i&1) == 0:
i += 1
result = min(result, i)
if (zero&1) == 0:
i = max(ceil_divide(zero, k), ceil_divide(zero, len(s)-k))
if (i&1) == 1:
i += 1
result = min(result, i)
return result if result != float("inf") else -1
# Time: O(n)
# Space: O(1)
# math
class Solution2(object):
def minOperations(self, s, k):
"""
:type s: str
:type k: int
:rtype: int
"""
def ceil_divide(a, b):
return (a+b-1)//b
zero = s.count('0')
if len(s) == k:
return 0 if zero == 0 else 1 if zero == len(s) else -1
result = float("inf")
i = max(ceil_divide(zero, k), ceil_divide(len(s)-zero, len(s)-k))
if (i&1) == 0:
i += 1
if ((i*k-zero)&1) == 0: # (k&1) == (zero&1)
result = min(result, i)
i = max(ceil_divide(zero, k), ceil_divide(zero, len(s)-k))
if (i&1) == 1:
i += 1
if ((i*k-zero)&1) == 0: # (zero&1) == 0
result = min(result, i)
return result if result != float("inf") else -1
# Time: O(n)
# Space: O(1)
# math
class Solution3(object):
def minOperations(self, s, k):
"""
:type s: str
:type k: int
:rtype: int
"""
zero = s.count('0')
for i in xrange(len(s)+1):
if (i*k-zero)&1:
continue
if i&1:
if zero <= i*k <= zero*i+(len(s)-zero)*(i-1):
return i
else:
if zero <= i*k <= zero*(i-1)+(len(s)-zero)*i:
return i
return -1