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 (aORb==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 <= 10^91 <= b <= 10^91 <= c <= 10^9
struct Solution;
impl Solution {
fn min_flips(a: i32, b: i32, c: i32) -> i32 {
let mut res = 0;
for i in 0..32 {
let aa = (a >> i) & 1;
let bb = (b >> i) & 1;
let cc = (c >> i) & 1;
if cc == 0 {
if aa == 1 {
res += 1;
}
if bb == 1 {
res += 1;
}
} else {
if aa == 0 && bb == 0 {
res += 1;
}
}
}
res
}
}
#[test]
fn test() {
let a = 2;
let b = 6;
let c = 5;
let res = 3;
assert_eq!(Solution::min_flips(a, b, c), res);
let a = 4;
let b = 2;
let c = 7;
let res = 1;
assert_eq!(Solution::min_flips(a, b, c), res);
let a = 1;
let b = 2;
let c = 3;
let res = 0;
assert_eq!(Solution::min_flips(a, b, c), res);
let a = 8;
let b = 3;
let c = 5;
let res = 3;
assert_eq!(Solution::min_flips(a, b, c), res);
}