Introduction

In this blog post, we will explore how to solve the LeetCode problem "3209" 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 array of integers nums and an integer k, return the number of subarrays of nums where the bitwise AND of the elements of the subarray equals k. Example 1: Input: nums = [1,1,1], k = 1 Output: 6 Explanation: All subarrays contain only 1's. Example 2: Input: nums = [1,1,2], k = 1 Output: 3 Explanation: Subarrays having an AND value of 1 are: [1,1,2], [1,1,2], [1,1,2]. Example 3: Input: nums = [1,2,3], k = 2 Output: 2 Explanation: Subarrays having an AND value of 2 are: [1,2,3], [1,2,3]. Constraints: 1 <= nums.length <= 105 0 <= nums[i], k <= 109

Explanation

Here's the solution:

• Iterate and Calculate AND: Iterate through the array using nested loops. The outer loop selects the starting index of the subarray, and the inner loop expands the subarray to the right. Calculate the bitwise AND of the current subarray elements.
• Count Matching Subarrays: If the calculated bitwise AND is equal to k, increment the count.

• Optimization: Since the bitwise AND is monotonically decreasing, we can break the inner loop early if the current AND value is less than k, as it will only decrease further.

• Runtime Complexity: O(n^2), Storage Complexity: O(1)

Code

    def count_subarrays(nums, k):
    """
    Counts the number of subarrays of nums where the bitwise AND of the elements of the subarray equals k.

    Args:
        nums: A list of integers.
        k: An integer.

    Returns:
        The number of subarrays of nums where the bitwise AND of the elements of the subarray equals k.
    """
    count = 0
    n = len(nums)

    for i in range(n):
        current_and = nums[i]
        if current_and == k:
            count += 1

        for j in range(i + 1, n):
            current_and &= nums[j]
            if current_and == k:
                count += 1
            elif current_and < k:
                break

    return count