How do you solve Number of Subarrays With AND Value of K on LeetCode?

Number of Subarrays With AND Value of K (LeetCode #3209) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The AND operation can only reduce or maintain the value, never increase it.

What is the brute force solution for Number of Subarrays With AND Value of K?

The brute force approach for Number of Subarrays With AND Value of K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray and calculating the bitwise AND for each. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Subarrays With AND Value of K?

Number of Subarrays With AND Value of K uses the following concepts: Array, Binary Search, Bit Manipulation, Segment Tree. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Subarrays With AND Value of K?

Always explain your thought process and the properties of operations like AND. Consider edge cases, such as when k is larger than any possible AND value. Practice identifying patterns in problems to recognize when to apply specific techniques.

What are common mistakes when solving Number of Subarrays With AND Value of K?

Not considering the properties of the AND operation, leading to unnecessary calculations. Failing to reset the AND value for each new starting index.

#3209

Number of Subarrays With AND Value of K

Hard

ArrayBinary SearchBit ManipulationSegment TreeHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a sliding window approach combined with bit manipulation to efficiently count valid subarrays. This reduces unnecessary calculations and leverages properties of the AND operation.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable to count valid subarrays and a pointer for the right end of the window.
2Step 2: Iterate through the array with a left pointer, maintaining a right pointer to extend the window.
3Step 3: For each left pointer position, reset the AND value and move the right pointer to find valid subarrays.
4Step 4: If the AND value equals k, count all subarrays ending at the right pointer.
5Step 5: Move the left pointer and repeat until all positions are checked.

solution.py14 lines

1def countSubarrays(nums, k):
2    count = 0
3    n = len(nums)
4    for left in range(n):
5        and_value = nums[left]
6        if and_value == k:
7            count += 1
8        for right in range(left + 1, n):
9            and_value &= nums[right]
10            if and_value == k:
11                count += 1
12            if and_value < k:
13                break
14    return count

ℹ

Complexity note: The time complexity is O(n) because we efficiently traverse the array while maintaining a running AND value. The space complexity is O(1) as we only use a few variables for counting.

1The AND operation can only reduce or maintain the value, never increase it.
2If the AND value drops below k, further elements will not help, allowing us to break early.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.