Problem Description

Given an array of integers nums and an integer k, you are tasked with finding the number of subarrays of nums where the bitwise AND of the elements of the subarray equals k. A subarray is defined as a contiguous section of the array.

Intuition

The solution revolves around dynamically calculating possible results of bitwise AND operations for subarrays. This is achieved by fixing a right endpoint r and iterating through possible left endpoints l in this manner:

1. Hash Table + Enumeration: First, fix the right endpoint r and iterate over possible left endpoints l. For each potential subarray ending at r, calculate the bitwise AND of the subarray and maintain a count of different outcomes using a hash table (Counter in Python).

2. Dynamic Maintenance: As the right endpoint extends (i.e., r increases), update possible results by computing the bitwise AND of each previous observed result with nums[r]. This accounts for new subarrays that end at the new r.

3. Count Occurrences: Each time a subarray with a bitwise AND result matching k is found, increase the count. Different subarray combinations are considered as new elements are added to the tracking collection.

By dynamically maintaining possible results and their counts and extending each subarray step-by-step, the solution efficiently counts valid subarrays in linear time complexity relative to the array length, taking advantage of the property that bitwise AND results in decreasing values as more elements are included in the subarray.

Learn more about Segment Tree and Binary Search patterns.

Solution Approach

The approach leverages a combination of hash tables and iterative enumeration to efficiently count subarrays with a bitwise AND result equal to k. Here's how it works:

1. Initialization:

   • Use a Counter, named pre, to store the frequency of each possible result from the bitwise AND operation as subarrays are considered. The Counter helps in maintaining counts of specific numbers rapidly.
   • Initialize an ans variable to zero, which will hold the final count of valid subarrays.

2. Iterate through Array Elements:

   • For each element x in nums, initialize a new Counter, cur, to store the results of calculations involving x.

3. Calculate Bitwise AND Results:

   • For each previous result, y, in the pre Counter, compute x & y and update the cur Counter with this new result and its associated count. This step effectively extends all subarrays that previously gave the result y to include the new element x.
   • Additionally, add x itself to cur with a count of 1 since x can be the start of new subarrays.

4. Update the Answer:

   • Add to ans the count of times k appears in the cur Counter. This step ensures that all subarrays ending at the current element x and having a bitwise AND equal to k are counted.

5. Prepare for Next Iteration:

   • Assign cur to pre for the next iteration. This transfer means that all potential subarrays ending right before moving to the next element are accessed efficiently.

Using this method ensures that as each new element is processed, all possible preceding results are taken into account. This makes the implementation time-efficient for the problem's constraints:

class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        ans = 0
        pre = Counter()  # This willcount the occurrences of specific AND results
        for x in nums:
            cur = Counter()  # For storing AND results that finish with the current element
            for y, v in pre.items():
                cur[x & y] += v  # Extend previous results with the current element
            cur[x] += 1  # Consider subarray consisting of the current element itself
            ans += cur[k]  # Add to answer count of necessary results found
            pre = cur  # Move on to the next iteration with updated results
        return ans

This strategic use of counting through hash tables and bitwise operations provides a comprehensive and efficient solution to the problem.

Example Walkthrough

Let's walk through a simple example to understand the solution approach.

Consider the array nums = [2, 1, 3] and k = 1. We need to find subarrays whose bitwise AND equals 1.

1. Initialization:

   • Start with ans = 0 indicating the count of valid subarrays.
   • A pre Counter is initialized to keep track of previous results. Initially, it's empty.

2. Processing 2:

   • x = 2, initialize cur = Counter().
   • Calculate x & y for each result in pre. Since pre is empty, this step is skipped.
   • Add 2 & 2 to cur, i.e., cur[2] = 1.
   • No subarray matches k yet, so ans remains 0.
   • Set pre = cur, so now pre = Counter({2: 1}).

3. Processing 1:

   • x = 1, initialize cur = Counter().
   • Calculate x & y for each result in pre: 1 & 2 = 0, so cur[0] = 1.
   • Add 1 & 1 to cur, i.e., cur[1] = 1.
   • cur has 1 (matching k=1) once, so increment ans by 1. Now, ans = 1.
   • Set pre = cur, so pre = Counter({0: 1, 1: 1}).

4. Processing 3:

   • x = 3, initialize cur = Counter().
   • Calculate x & y for each result in pre:
     • 3 & 0 = 0, update cur[0] = 1.
     • 3 & 1 = 1, update cur[1] = 1.
   • Add 3 & 3 = 3 to cur, i.e., cur[3] = 1.
   • cur has 1 again (matching k=1), so increment ans by 1. Now, ans = 2.
   • Prepare for next element (though the array ends here), by setting pre = cur.

At the end of the iteration, ans = 2, indicating there are 2 subarrays ([1] and [1, 3]) where the bitwise AND equals k=1.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n * log M). Here's why:

• The outer loop runs n times, where n is the length of the nums array.
• For each element x in nums, the inner loop iterates over each entry y, v in the pre dictionary, where pre is a Counter of previously computed bitwise AND values.
• The size of pre can be proportional to log M due to the nature of bitwise operations, where M is the maximum value in nums.
• The overall computation in the inner loop is limited to log M operations for each element due to the constraints of bitwise values.

The space complexity of the code is O(log M). Here's why:

• The pre and cur dictionaries store combinations of bitwise AND results.
• The space used by these dictionaries grows with the distinct bitwise AND results, which is limited by log M due to bitwise operation constraints.

Learn more about how to find time and space complexity quickly.