README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0400-0499/0476.Number%20Complement/README_EN.md	Bit Manipulation

476. Number Complement

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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);
}