Write a method that finds the maximum of two numbers. You should not use if-else or any other comparison operator.
Example:
Input: a = 1, b = 2
Output: 2
We can extract the sign bit
Then the final result is
The time complexity is
class Solution:
def maximum(self, a: int, b: int) -> int:
k = (int(((a - b) & 0xFFFFFFFFFFFFFFFF) >> 63)) & 1
return a * (k ^ 1) + b * k
class Solution {
public int maximum(int a, int b) {
int k = (int) (((long) a - (long) b) >> 63) & 1;
return a * (k ^ 1) + b * k;
}
}
class Solution {
public:
int maximum(int a, int b) {
int k = ((static_cast<long long>(a) - static_cast<long long>(b)) >> 63) & 1;
return a * (k ^ 1) + b * k;
}
};
func maximum(a int, b int) int {
k := (a - b) >> 63 & 1
return a*(k^1) + b*k
}
function maximum(a: number, b: number): number {
const k: number = Number(((BigInt(a) - BigInt(b)) >> BigInt(63)) & BigInt(1));
return a * (k ^ 1) + b * k;
}