Leetcode #2166: Design Bitset
In this guide, we solve Leetcode #2166 Design Bitset in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
A Bitset is a data structure that compactly stores bits. Implement the Bitset class: Bitset(int size) Initializes the Bitset with size bits, all of which are 0.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Array, Hash Table, String
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input
["Bitset", "fix", "fix", "flip", "all", "unfix", "flip", "one", "unfix", "count", "toString"]
[[5], [3], [1], [], [], [0], [], [], [0], [], []]
Output
[null, null, null, null, false, null, null, true, null, 2, "01010"]
Explanation
Bitset bs = new Bitset(5); // bitset = "00000".
bs.fix(3); // the value at idx = 3 is updated to 1, so bitset = "00010".
bs.fix(1); // the value at idx = 1 is updated to 1, so bitset = "01010".
bs.flip(); // the value of each bit is flipped, so bitset = "10101".
bs.all(); // return False, as not all values of the bitset are 1.
bs.unfix(0); // the value at idx = 0 is updated to 0, so bitset = "00101".
bs.flip(); // the value of each bit is flipped, so bitset = "11010".
bs.one(); // return True, as there is at least 1 index with value 1.
bs.unfix(0); // the value at idx = 0 is updated to 0, so bitset = "01010".
bs.count(); // return 2, as there are 2 bits with value 1.
bs.toString(); // return "01010", which is the composition of bitset.
Python Solution
class Bitset:
def __init__(self, size: int):
self.a = ['0'] * size
self.b = ['1'] * size
self.cnt = 0
def fix(self, idx: int) -> None:
if self.a[idx] == '0':
self.a[idx] = '1'
self.cnt += 1
self.b[idx] = '0'
def unfix(self, idx: int) -> None:
if self.a[idx] == '1':
self.a[idx] = '0'
self.cnt -= 1
self.b[idx] = '1'
def flip(self) -> None:
self.a, self.b = self.b, self.a
self.cnt = len(self.a) - self.cnt
def all(self) -> bool:
return self.cnt == len(self.a)
def one(self) -> bool:
return self.cnt > 0
def count(self) -> int:
return self.cnt
def toString(self) -> str:
return ''.join(self.a)
# Your Bitset object will be instantiated and called as such:
# obj = Bitset(size)
# obj.fix(idx)
# obj.unfix(idx)
# obj.flip()
# param_4 = obj.all()
# param_5 = obj.one()
# param_6 = obj.count()
# param_7 = obj.toString()
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.