476. Number Complement 

Description

The complement of an integer is the integer you get when you flip all the 0's to 1's and all the 1's to 0's in its binary representation.

For example, The integer 5 is "101" in binary and its complement is "010" which is the integer 2.

Given an integer num, return its complement.

Example 1:

Input: num = 5
Output: 2
Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.

Example 2:

Input: num = 1
Output: 0
Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0.

Constraints:

1 <= num < 2³¹

Note: This question is the same as 1009: https://leetcode.com/problems/complement-of-base-10-integer/

Solutions

Solution 1: Bit Manipulation

According to the problem description, we can use XOR operation to implement the flipping operation, the steps are as follows:

First, we find the highest bit of $1$ in the binary representation of $\textit{num}$, and the position is denoted as $k$.

Then, we construct a binary number, where the $k$-th bit is $0$ and the rest of the lower bits are $1$, which is $2^k - 1$;

Finally, we perform XOR operation on $\textit{num}$ and the constructed binary number to get the answer.

The time complexity is $O(\log \textit{num})$, where $\textit{num}$ is the input integer. The space complexity is $O(1)$.

Python3

class Solution:
    def findComplement(self, num: int) -> int:
        return num ^ ((1 << num.bit_length()) - 1)

Java

class Solution {
    public int findComplement(int num) {
        return num ^ ((1 << (32 - Integer.numberOfLeadingZeros(num))) - 1);
    }
}

C++

class Solution {
public:
    int findComplement(int num) {
        return num ^ ((1LL << (64 - __builtin_clzll(num))) - 1);
    }
};

Go

func findComplement(num int) int {
	return num ^ ((1 << bits.Len(uint(num))) - 1)
}

TypeScript

function findComplement(num: number): number {
    return num ^ (2 ** num.toString(2).length - 1);
}

JavaScript

/**
 * @param {number} num
 * @return {number}
 */
var findComplement = function (num) {
    return num ^ (2 ** num.toString(2).length - 1);
};

Solution 2: Bit Manipulation. Inversion + AND

TypeScript

function findComplement(num: number): number {
    return ~num & (2 ** num.toString(2).length - 1);
}

JavaScript

/**
 * @param {number} num
 * @return {number}
 */
function findComplement(num) {
    return ~num & (2 ** num.toString(2).length - 1);
}