Introduction

In this blog post, we will explore how to solve the LeetCode problem "2166" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. void fix(int idx) Updates the value of the bit at the index idx to 1. If the value was already 1, no change occurs. void unfix(int idx) Updates the value of the bit at the index idx to 0. If the value was already 0, no change occurs. void flip() Flips the values of each bit in the Bitset. In other words, all bits with value 0 will now have value 1 and vice versa. boolean all() Checks if the value of each bit in the Bitset is 1. Returns true if it satisfies the condition, false otherwise. boolean one() Checks if there is at least one bit in the Bitset with value 1. Returns true if it satisfies the condition, false otherwise. int count() Returns the total number of bits in the Bitset which have value 1. String toString() Returns the current composition of the Bitset. Note that in the resultant string, the character at the ith index should coincide with the value at the ith bit of the Bitset. Example 1: 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. Constraints: 1 <= size <= 105 0 <= idx <= size - 1 At most 105 calls will be made in total to fix, unfix, flip, all, one, count, and toString. At least one call will be made to all, one, count, or toString. At most 5 calls will be made to toString.

Explanation

• Use a boolean array for bit storage: This allows for direct bit manipulation (setting, unsetting, and flipping).
  • Maintain a flip counter: Instead of physically flipping the bits during the flip operation, increment a counter. The value of a bit is then determined by combining its stored value and the flip counter's parity.
  • Keep track of the number of ones: Increment or decrement a count variable when bits are fixed or unfixed, considering the effect of the flip counter.

• Runtime Complexity: O(1) for fix, unfix, flip, all, one, and count. O(n) for toString where n is the size of the Bitset. Storage Complexity: O(n) where n is the size of the Bitset.

Code

    class Bitset:

    def __init__(self, size: int):
        self.size = size
        self.bits = [False] * size
        self.flip_count = 0
        self.ones = 0

    def fix(self, idx: int) -> None:
        actual_value = self.bits[idx] ^ (self.flip_count % 2 == 1)
        if not actual_value:
            self.bits[idx] = True
            self.ones += 1

    def unfix(self, idx: int) -> None:
        actual_value = self.bits[idx] ^ (self.flip_count % 2 == 1)
        if actual_value:
            self.bits[idx] = False
            self.ones -= 1

    def flip(self) -> None:
        self.flip_count += 1
        self.ones = self.size - self.ones

    def all(self) -> bool:
        return self.ones == self.size

    def one(self) -> bool:
        return self.ones > 0

    def count(self) -> int:
        return self.ones

    def toString(self) -> str:
        result = ""
        for i in range(self.size):
            result += "1" if self.bits[i] ^ (self.flip_count % 2 == 1) else "0"
        return result