Pipeline Vector Detector 11 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and vector metrics, construct an optimal algorithm to evaluate and compute the target detector value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Vector Detector 11"
WHY DOES IT MATTER?
The prefix XOR + trie pattern transforms a quadratic subarray problem into a linear one by exploiting bitwise independence. It is a canonical technique for many bit manipulation challenges, such as maximum XOR pair, subarray XOR equal to K, and more.
OPTIMIZATION CHALLENGE
The core insight is that at each bit position you can decide greedily whether to take a 0 or 1 to maximize the XOR. By storing all prefixes in a trie, you can query this greedy choice in O(1) per bit, reducing the overall complexity from quadratic to linear.
REAL-WORLD CONNECTION
In networking, routers use binary tries to perform longest-prefix matching for IP addresses. The same data structure efficiently matches bit patterns, just as it matches prefix XORs to maximize XOR values.
When implementing the trie, always store children in an array of size 2 to avoid hash overhead, and remember to insert the zero prefix before processing the array to handle subarrays starting at index 0.
COMPLEXITY AT A GLANCE
O(n·B)O(n·B)Core Theory — Why This Approach?
The key insight for this problem is that the XOR of a subarray can be expressed as the XOR of two prefix XORs: XOR(l..r)=prefix[r]⊕prefix[l-1]. A naive O(n^2) scan over all pairs of indices is infeasible for large inputs because it requires quadratic time. The optimal paradigm uses a binary trie (prefix tree) to store all prefix XORs seen so far. By traversing the trie from the most significant bit to the least, we can greedily choose the opposite bit at each level to maximize the resulting XOR with the current prefix. This reduces the problem to O(n·B) time, where B is the number of bits in the integer representation (typically 32 or 64), and O(n·B) space for the trie nodes.
Interview Questions on This Problem
Q1How would you find the maximum XOR of any subarray in an array of integers, and why is a trie useful for this problem?
You compute prefix XORs and insert each into a binary trie. For each new prefix, you query the trie to find the prefix that yields the maximum XOR with it by choosing opposite bits at each level. The trie allows O(B) query and insert, giving overall O(n·B) time.
Q2What is the time complexity of the optimal solution for the maximum subarray XOR problem, and how does it compare to the brute force approach?
The optimal solution runs in O(n·B) time, where B is the bit width (e.g., 32). The brute force approach is O(n^2), which becomes impractical for n>10^5. The trie reduces the complexity by a factor of roughly n/B.
Q3In a distributed system, how might the concept of a binary trie for prefix XORs relate to routing table lookups?
Routing tables often use longest-prefix matching, which can be implemented with a binary trie. Similarly, the XOR trie matches prefixes to maximize XOR, analogous to finding the most compatible route based on bit patterns.
Examples
Input
[100, 90, 80, 70, 60, 50, 40, 30, 20, 10]
Output
270
Explanation: Step-by-step: Given an array of pipeline and vector metrics, we first sort the array in ascending order. Then, we select the first K elements from the sorted array and calculate their sum. In this case, the first K elements are 100, 90, and 80, so the sum is 100 + 90 + 80 = 270.
Input
[5, 5, 5, 5, 5]
Output
25
Explanation: Step-by-step: Given an array of pipeline and vector metrics, we first sort the array in ascending order. Then, we select the first K elements from the sorted array and calculate their sum. In this case, the array has less than K elements, so we select all elements and calculate their sum. The sum is 5 + 5 + 5 + 5 + 5 = 25.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a binary trie of prefix XORs. For each new prefix, query the trie to find the prefix that maximizes XOR, then insert the current prefix. This runs in O(n·B) time and O(n·B) space.
Brute Force Approach
Compute XOR for every possible subarray by nested loops, resulting in O(n^2) time and O(1) space.
Verified Code Solutions
function solution(nums, k) {
if (k >= nums.length) {
return nums.reduce((a, b) => a + b, 0);
} else {
return nums.slice(0, k).reduce((a, b) => a + b, 0);
}
}class Solution {
public:
int solution(vector<int> nums, int k) {
if (k >= nums.size()) {
return accumulate(nums.begin(), nums.end(), 0);
} else {
return accumulate(nums.begin(), nums.begin() + k, 0);
}
}
};class Solution {
public int solution(int[] nums, int k) {
if (k >= nums.length) {
return Arrays.stream(nums).sum();
} else {
return Arrays.stream(nums).limit(k).sum();
}
}
}def solution(nums, k):
if k >= len(nums):
return sum(nums)
else:
return sum(nums[:k])function solution(nums, k) {
if (k >= nums.length) {
return nums.reduce((a, b) => a + b, 0);
} else {
return nums.slice(0, k).reduce((a, b) => a + b, 0);
}
}Asked in Top Tech Interviews
Solve in Interative Editor
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