2429. Minimize XOR
Description
Given two positive integers num1
and num2
, find the positive integer x
such that:
x
has the same number of set bits asnum2
, and- The value
x XOR num1
is minimal.
Note that XOR
is the bitwise XOR operation.
Return the integer x
. The test cases are generated such that x
is uniquely determined.
The number of set bits of an integer is the number of 1
's in its binary representation.
Example 1:
Input: num1 = 3, num2 = 5 Output: 3 Explanation: The binary representations of num1 and num2 are 0011 and 0101, respectively. The integer 3 has the same number of set bits as num2, and the value 3 XOR 3 = 0 is minimal.
Example 2:
Input: num1 = 1, num2 = 12 Output: 3 Explanation: The binary representations of num1 and num2 are 0001 and 1100, respectively. The integer 3 has the same number of set bits as num2, and the value 3 XOR 1 = 2 is minimal.
Constraints:
1 <= num1, num2 <= 109
Solutions
Solution 1: Greedy + Bit Manipulation
According to the problem description, we first calculate the number of set bits $cnt$ in $num2$, then enumerate each bit of $num1$ from high to low. If the bit is $1$, we set the corresponding bit in $x$ to $1$ and decrement $cnt$ by $1$, until $cnt$ is $0$. If $cnt$ is still not $0$ at this point, we start from the low bit and set each bit of $num1$ that is $0$ to $1$, and decrement $cnt$ by $1$, until $cnt$ is $0$.
The time complexity is $O(\log n)$, and the space complexity is $O(1)$. Here, $n$ is the maximum value of $num1$ and $num2$.
1 2 3 4 5 6 7 8 9 10 11 12 13 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 |
|