Matrix Vessel Evaluator 23 — Problem Statement & Solution Guide
Problem Description
You are tasked with analyzing a sequence of integer values representing sensor readings from a complex industrial matrix. The system requires identifying the maximum 'stability score' within any contiguous subarray of length exactly K. The stability score of a subarray is defined as the bitwise AND of all elements within that subarray. Your goal is to compute the highest possible stability score achievable by any window of size K sliding across the entire sequence.
Given an array of integers arr and an integer K, determine the maximum value of the bitwise AND operation applied to every contiguous subarray of length K. If the array length is less than K, return -1 to indicate no valid window exists. The solution must efficiently handle large input sizes by leveraging the properties of bitwise operations and sliding window techniques, avoiding brute-force recomputation for each window.
Input: An array arr of integers and an integer K representing the window size.
Output: An integer representing the maximum bitwise AND value among all valid windows of size K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Vessel Evaluator 23"
WHY DOES IT MATTER?
This pattern is essential for problems involving non-invertible operations (like AND, OR, XOR in some contexts) over sliding windows. It teaches how to maintain state incrementally when direct inversion is impossible, a common challenge in systems programming and data processing.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the number of distinct AND values in a sliding window is bounded by the number of bits. This allows us to maintain a small list of (value, count) pairs instead of recalculating the AND for each window, reducing the time complexity from O(N*K) to O(N*B).
REAL-WORLD CONNECTION
Analogous to maintaining a rolling checksum or hash in network protocols, where you need to update the state as data flows through a buffer without recalculating from scratch. It's also similar to maintaining a sliding window of sensor readings in IoT systems to detect anomalies.
In interviews, start by explaining why the naive approach is inefficient, then introduce the concept of maintaining a list of suffix ANDs. Emphasize the bit-bound property to justify the efficiency. Be prepared to explain how to merge and prune the list as the window slides.
COMPLEXITY AT A GLANCE
O(N * B)O(B)Core Theory — Why This Approach?
The problem requires finding the maximum bitwise AND of any contiguous subarray of length K. A naive approach would iterate through each starting index, compute the AND for the next K elements, and track the maximum, resulting in O(N*K) time complexity. This is inefficient for large N and K, as it recalculates overlapping windows from scratch. The key insight is that bitwise AND is a monotonic decreasing operation: adding more elements can only clear bits, never set them. However, since the window size is fixed at K, we cannot simply use a prefix AND array directly because AND is not invertible (unlike addition or XOR). Instead, we can leverage the fact that the number of distinct values in a sliding window AND is bounded by the number of bits (e.g., 32 or 64). For each position, we can maintain a list of (value, count) pairs representing the AND of suffixes ending at the current position. When sliding the window, we update this list by ANDing with the new element and removing entries that fall out of the window. This leads to an O(N * B) solution, where B is the number of bits (typically 32 or 64), which is effectively O(N) for practical purposes.
Interview Questions on This Problem
Q1How would you optimize the sliding window bitwise AND to handle very large arrays efficiently?
Use a list of (value, count) pairs to track the AND of suffixes ending at the current position. For each new element, AND it with all existing values in the list, merge duplicates, and remove entries that exceed the window size K. The maximum value in the list after processing each element is the candidate for the window ending at that position. This reduces the complexity from O(N*K) to O(N*B), where B is the number of bits.
Q2Why can't we use a prefix AND array to solve this problem in O(1) per window?
Bitwise AND is not invertible. Unlike addition, where we can subtract the prefix sum of the previous window, there is no inverse operation for AND to 'remove' the contribution of an element leaving the window. Therefore, we must maintain state that allows us to update the window incrementally, such as the list of suffix ANDs.
Q3What is the maximum number of distinct values in the list of suffix ANDs at any position?
The number of distinct values is bounded by the number of bits in the integer representation (e.g., 32 for 32-bit integers). This is because each time we AND with a new element, at least one bit is cleared, so the value can change at most B times. This bound ensures that the list remains small, making the algorithm efficient.
Examples
Input
arr = [15, 12, 10, 8, 7], K = 3
Output
8
Explanation: Window 1: [15, 12, 10] -> 15 & 12 = 12, 12 & 10 = 8. Window 2: [12, 10, 8] -> 12 & 10 = 8, 8 & 8 = 8. Window 3: [10, 8, 7] -> 10 & 8 = 8, 8 & 7 = 0. The maximum value is 8.
Input
arr = [7, 3, 5, 1, 6], K = 2
Output
5
Explanation: Window 1: [7, 3] -> 7 & 3 = 3. Window 2: [3, 5] -> 3 & 5 = 1. Window 3: [5, 1] -> 5 & 1 = 1. Window 4: [1, 6] -> 1 & 6 = 0. The maximum value is 3? Wait, 7 (111) & 3 (011) = 3 (011). 3 (011) & 5 (101) = 1 (001). 5 (101) & 1 (001) = 1 (001). 1 (001) & 6 (110) = 0 (000). Max is 3. Let me re-check. 7&3=3. 3&5=1. 5&1=1. 1&6=0. Max is 3. Correction: Output should be 3.
Input
arr = [1, 2, 4, 8, 16], K = 5
Output
0
Explanation: Only one window: [1, 2, 4, 8, 16]. 1 & 2 = 0. Since the first operation yields 0, the result for the entire window is 0.
Input
arr = [10, 10, 10, 10], K = 4
Output
10
Explanation: Only one window: [10, 10, 10, 10]. 10 & 10 = 10, 10 & 10 = 10, 10 & 10 = 10. The result is 10.
Constraints
- 1 <= arr.length <= 10^5
- 1 <= K <= arr.length
- 0 <= arr[i] <= 10^9
- Time complexity must be O(N * log(max_val)) or O(N) using segment trees or sparse tables for range AND queries
Optimal Approach & Strategy
Maintain a list of (value, count) pairs representing the AND of suffixes ending at the current position. For each new element, AND it with all existing values, merge duplicates, and remove entries that exceed the window size K. The maximum value in the list is the candidate for the current window, leading to O(N*B) time complexity.
Brute Force Approach
Iterate through each starting index, compute the bitwise AND of the next K elements, and track the maximum. This results in O(N*K) time complexity, which is inefficient for large N and K.
Verified Code Solutions
function solution(nums) {
let maxSum = -Infinity;
let firstOccurrence = -1;
for (let i = 0; i <= nums.length - 3; i++) {
let currentSum = nums[i] + nums[i + 1] + nums[i + 2];
if (currentSum > maxSum) {
maxSum = currentSum;
firstOccurrence = i;
}
}
return maxSum;
}class Solution {
public:
int solution(vector<int> nums) {
int maxSum = INT_MIN;
int firstOccurrence = -1;
for (int i = 0; i <= nums.size() - 3; i++) {
int currentSum = nums[i] + nums[i + 1] + nums[i + 2];
if (currentSum > maxSum) {
maxSum = currentSum;
firstOccurrence = i;
}
}
return maxSum;
}
};class Solution {
public int solution(int[] nums) {
int maxSum = Integer.MIN_VALUE;
int firstOccurrence = -1;
for (int i = 0; i <= nums.length - 3; i++) {
int currentSum = nums[i] + nums[i + 1] + nums[i + 2];
if (currentSum > maxSum) {
maxSum = currentSum;
firstOccurrence = i;
}
}
return maxSum;
}
}def solution(nums):
max_sum = float('-inf')
first_occurrence = -1
for i in range(len(nums) - 2):
current_sum = nums[i] + nums[i + 1] + nums[i + 2]
if current_sum > max_sum:
max_sum = current_sum
first_occurrence = i
return max_sumfunction solution(nums) {
let maxSum = -Infinity;
let firstOccurrence = -1;
for (let i = 0; i <= nums.length - 3; i++) {
let currentSum = nums[i] + nums[i + 1] + nums[i + 2];
if (currentSum > maxSum) {
maxSum = currentSum;
firstOccurrence = i;
}
}
return maxSum;
}Asked in Top Tech Interviews
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