1. Problem
1318. Minimum Flips to Make a OR b Equal to c
Given 3 positives numbers a, b and c. Return the minimum flips required in some bits of a and b to make ( a OR b == c ). (bitwise OR operation). Flip operation consists of change any single bit 1 to 0 or change the bit 0 to 1 in their binary representation.
Example 1:
Input: a = 2, b = 6, c = 5
Output: 3
Explanation: After flips a = 1 , b = 4 , c = 5 such that (a OR b == c)
Example 2:
Input: a = 4, b = 2, c = 7
Output: 1
Example 3:
Input: a = 1, b = 2, c = 3
Output: 0
Constraints:
 1 <= a <= 10^9
 1 <= b <= 10^9
 1 <= c <= 10^9
2. Solution
I solved this problem like this.
 Step

Get (a b) and XOR C. If (a b) ^ c bit 1, we have to change a or b.  If bit is 0, we don’t need to check that index.
 If bit is 1,
 If c is 1, a & b are both 0. We just change 1 bit.
 If c is 0, a & b are either [(0, 1), (1, 0), (1, 1)]. We change bit and if bit is 1, we change to 0.

class Solution:
def minFlips(self, a: int, b: int, c: int) > int:
answer = 0
sub = (a  b) ^ c
size = len(bin(a  b  c))  2
for i in range(size):
x = bin(sub)[2:].zfill(size)[i]
if x != '1':
continue
if bin(c)[2:].zfill(size)[i] == '1':
answer += 1
else:
answer += int(bin(a)[2:].zfill(size)[i]) + int(bin(b)[2:].zfill(size)[i])
return answer
Other solution is more better.
class Solution:
def minFlips(self, a: int, b: int, c: int) > int:
return ((a  b) ^ c).bit_count() + (a & b & ((a  b) ^ c)).bit_count()