LeetCode Problem Workspace
Complement of Base 10 Integer
In this problem, you need to return the complement of a given integer by flipping its binary digits.
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Code langs
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Practice Focus
Easy · Bit Manipulation-driven solution strategy
Answer-first summary
In this problem, you need to return the complement of a given integer by flipping its binary digits.
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This problem requires flipping the binary digits of an integer to get its complement. The solution can be approached through bitwise operations, leveraging the fact that a binary number plus its complement equals a series of ones. The challenge lies in efficiently computing the result, especially for large values of n.
Problem Statement
The complement of an integer is derived by flipping all the bits in its binary representation, changing 1's to 0's and 0's to 1's.
Given an integer n, you are tasked with returning its complement, where the complement is calculated by flipping each bit in its binary form.
Examples
Example 1
Input: n = 5
Output: 2
5 is "101" in binary, with complement "010" in binary, which is 2 in base-10.
Example 2
Input: n = 7
Output: 0
7 is "111" in binary, with complement "000" in binary, which is 0 in base-10.
Example 3
Input: n = 10
Output: 5
10 is "1010" in binary, with complement "0101" in binary, which is 5 in base-10.
Constraints
- 0 <= n < 109
Solution Approach
Bitwise XOR approach
One efficient way to compute the complement is by using the XOR operation. First, calculate a mask where all bits are 1 for the number of bits in n. Then XOR the number with the mask, flipping all bits to produce the complement.
Inverting bits with subtraction
Another approach involves subtracting n from a number that consists entirely of 1's, which has the same bit-length as n. This subtraction inverts the bits of n, effectively providing the complement.
Handling edge cases
In cases where n equals 0, the complement is simply 1. This corner case needs special handling to ensure the solution accounts for it.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity of the solution depends on the number of bits in the integer n, which is O(log n). The space complexity is O(1), as the approach uses a constant amount of extra space.
What Interviewers Usually Probe
- Look for knowledge of bitwise operations and how they are used to manipulate binary representations.
- Assess understanding of how to handle edge cases such as when n is 0.
- Evaluate familiarity with time and space complexities related to bit manipulation problems.
Common Pitfalls or Variants
Common pitfalls
- Failing to account for the edge case where n equals 0, which should return 1.
- Incorrectly handling large values of n by not properly calculating the bit-length.
- Overcomplicating the problem by trying unnecessary operations rather than using simple bitwise manipulation.
Follow-up variants
- What if the complement is computed for a list of integers instead of a single value?
- Can the solution be optimized for extremely large values of n, such as those near 10^9?
- How can you apply this bitwise technique to other problems involving binary operations?
FAQ
How do you solve the 'Complement of Base 10 Integer' problem?
To solve this problem, use bitwise operations such as XOR to flip the binary digits of the given integer and return its complement.
What is the time complexity of the complement problem?
The time complexity is O(log n), as it depends on the number of bits in the integer n.
What is the space complexity for the complement problem?
The space complexity is O(1) because the solution uses a constant amount of extra space.
How should you handle edge cases in the complement problem?
When n equals 0, return 1 as its complement. Ensure to handle this edge case properly.
What is the pattern in the 'Complement of Base 10 Integer' problem?
The problem follows a bit manipulation pattern where the complement is computed by flipping the binary digits of an integer using bitwise operations.
Solution
Solution 1: Bit Manipulation
First, we check if $n$ is $0$. If it is, we return $1$.
class Solution:
def bitwiseComplement(self, n: int) -> int:
if n == 0:
return 1
ans = i = 0
while n:
ans |= ((n & 1 ^ 1) << i)
i += 1
n >>= 1
return ansContinue Topic
bit manipulation
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