How do you solve Design Bitset on LeetCode?

Design Bitset (LeetCode #2166) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a flip counter allows us to efficiently manage the state of bits without needing to traverse the entire array for every operation.

What is the brute force solution for Design Bitset?

The brute force approach for Design Bitset is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach initializes a list of bits and directly manipulates the bits based on the operations. It uses a simple list to store the bits, making it easy to understand but inefficient for larger sizes. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Design Bitset?

Design Bitset uses the following concepts: Array, Hash Table, String, Design. The recommended patterns to study are: Array, Hash Map.

What are the interview tips for Design Bitset?

Always clarify the requirements and constraints of the problem before diving into coding. Think about edge cases, such as what happens when all bits are flipped or when no bits are set. Explain your thought process clearly while coding; it helps the interviewer understand your approach.

What are common mistakes when solving Design Bitset?

Failing to account for the flip state when checking conditions or counting bits. Not considering the implications of multiple flips on the bit states.

#2166

Design Bitset

Medium

ArrayHash TableStringDesignArrayHash Map

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses an additional variable to track the number of flips and the state of the bits. This allows for efficient operations without needing to traverse the entire list for every operation.

⚙️

Algorithm

8 steps

1Step 1: Initialize a list of size 'size' with all elements set to 0 and a flip counter.
2Step 2: For 'fix(idx)', set the bit at index 'idx' to 1 if it is currently 0, and increment the flip counter if it was flipped.
3Step 3: For 'unfix(idx)', set the bit at index 'idx' to 0 if it is currently 1, and increment the flip counter if it was flipped.
4Step 4: For 'flip()', toggle the flip counter and the state of the bits.
5Step 5: For 'all()', check if the count of bits is equal to the size considering the flip state.
6Step 6: For 'one()', check if the count of bits is greater than 0 considering the flip state.
7Step 7: For 'count()', return the count of bits adjusted by the flip state.
8Step 8: For 'toString()', construct the string based on the current flip state.

solution.py28 lines

1class Bitset:
2    def __init__(self, size: int):
3        self.bits = [0] * size
4        self.flip_count = 0
5        self.size = size
6    
7    def fix(self, idx: int):
8        if self.bits[idx] == self.flip_count % 2:
9            self.bits[idx] = 1 - self.flip_count % 2
10    
11    def unfix(self, idx: int):
12        if self.bits[idx] != self.flip_count % 2:
13            self.bits[idx] = self.flip_count % 2
14    
15    def flip(self):
16        self.flip_count += 1
17    
18    def all(self) -> bool:
19        return self.count() == self.size
20    
21    def one(self) -> bool:
22        return self.count() > 0
23    
24    def count(self) -> int:
25        return sum(1 for b in self.bits if b == (self.flip_count % 2))
26    
27    def toString(self) -> str:
28        return ''.join(str(1 - (b ^ (self.flip_count % 2))) for b in self.bits)

ℹ

Complexity note: The time complexity is O(n) for the count and toString operations, while all other operations are O(1) due to the use of the flip counter.

1Using a flip counter allows us to efficiently manage the state of bits without needing to traverse the entire array for every operation.
2Understanding how to manipulate bits and their states can lead to more efficient algorithms.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.