Solving Leetcode Interviews in Seconds with AI: Count Number of Maximum Bitwise-OR Subsets
Introduction
In this blog post, we will explore how to solve the LeetCode problem "2044" 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
Given an integer array nums, find the maximum possible bitwise OR of a subset of nums and return the number of different non-empty subsets with the maximum bitwise OR. An array a is a subset of an array b if a can be obtained from b by deleting some (possibly zero) elements of b. Two subsets are considered different if the indices of the elements chosen are different. The bitwise OR of an array a is equal to a[0] OR a[1] OR ... OR a[a.length - 1] (0-indexed). Example 1: Input: nums = [3,1] Output: 2 Explanation: The maximum possible bitwise OR of a subset is 3. There are 2 subsets with a bitwise OR of 3: - [3] - [3,1] Example 2: Input: nums = [2,2,2] Output: 7 Explanation: All non-empty subsets of [2,2,2] have a bitwise OR of 2. There are 23 - 1 = 7 total subsets. Example 3: Input: nums = [3,2,1,5] Output: 6 Explanation: The maximum possible bitwise OR of a subset is 7. There are 6 subsets with a bitwise OR of 7: - [3,5] - [3,1,5] - [3,2,5] - [3,2,1,5] - [2,5] - [2,1,5] Constraints: 1 <= nums.length <= 16 1 <= nums[i] <= 105
Explanation
Here's the approach and Python code to solve this problem:
- Find the Maximum OR: First, calculate the maximum possible bitwise OR value that can be achieved from any subset of the input array
nums. Count Subsets with Maximum OR: Iterate through all possible non-empty subsets of
numsand count the number of subsets whose bitwise OR equals the maximum OR value found in the previous step.Runtime Complexity: O(n * 2n), where n is the length of
nums.- Storage Complexity: O(1)
Code
def count_max_or_subsets(nums):
"""
Finds the maximum possible bitwise OR of a subset of nums and returns the
number of different non-empty subsets with the maximum bitwise OR.
"""
max_or = 0
for num in nums:
max_or |= num
count = 0
n = len(nums)
for i in range(1, 1 << n): # Iterate through all non-empty subsets
current_or = 0
for j in range(n):
if (i >> j) & 1:
current_or |= nums[j]
if current_or == max_or:
count += 1
return count