Sensor Packet Synthesizer 45 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing sensor and packet metrics, construct an optimal algorithm to evaluate and compute the target synthesizer value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Packet Synthesizer 45"
WHY DOES IT MATTER?
Transforming sub‑array XOR queries into prefix operations collapses a quadratic problem into linear time.
OPTIMIZATION CHALLENGE
The key is to avoid recomputing XOR for overlapping ranges by reusing previously computed prefixes.
REAL-WORLD CONNECTION
Network packet checksum calculations use the same XOR‑folding technique to detect errors efficiently.
Initialize the prefix set with 0 before iteration; this captures sub‑arrays that start at index 0 and simplifies the logic.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
Bit‑manipulation problems often reduce to operations on binary representations, where XOR behaves like addition without carry. By maintaining a running prefix XOR, the XOR of any sub‑array can be expressed as the XOR of two prefixes, turning a quadratic search into a lookup problem. Naïve double loops recompute XOR for each interval, leading to O(n^2) time and quickly exhausting time limits for n up to 10^5. The optimal paradigm leverages a hash‑set (or trie) of previously seen prefixes, allowing each new element to be processed in O(1) amortized time while updating the best synthesizer value, thus achieving linear complexity.
Interview Questions on This Problem
Q1How does XOR help in computing sub‑array aggregates compared to sum?
XOR is its own inverse, so prefixXOR[i] ^ prefixXOR[j] yields the XOR of the sub‑array (i, j]. This property eliminates the need for subtraction and handles overlapping bits cleanly.
Q2Why is a hash‑set of prefix XORs sufficient for the maximum‑XOR sub‑array problem?
Each new prefix can be paired with any earlier prefix to form a candidate XOR; storing all earlier prefixes lets us test all possibilities in constant time per element. The set provides O(1) look‑ups without extra traversal.
Q3What edge case must be handled when the optimal synthesizer value is zero?
Zero can arise from an empty sub‑array or identical prefixes; the algorithm must initialize the answer with 0 and consider the case where no positive improvement is found. Explicitly checking the initial prefix (0) in the set avoids missing this scenario.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
35
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], we need to find the sum of all numbers greater than or equal to 5. We iterate through the array and add up all numbers that meet the condition. The correct sum is 1+2+3+4+5+6+7+8+9 = 45 is incorrect. The correct sum is 1+2+3+4+5 = 15 is incorrect. The correct sum is 5+6+7+8+9 = 35.
Input
[5, 6, 7, 8, 9, 10]
Output
35
Explanation: Step-by-step: Given the input array [5, 6, 7, 8, 9, 10], we need to find the sum of all numbers greater than or equal to 5. We iterate through the array and add up all numbers that meet the condition. The correct sum is 5+6+7+8+9 = 35.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a running prefix XOR and a hash‑set of all earlier prefixes; for each new element, compute candidate XORs in O(1) and update the answer, achieving O(n) time.
Brute Force Approach
Iterate over all possible start and end indices, compute XOR for each sub‑array, and keep the maximum; this is O(n^2) time.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] >= 5) {
sum += nums[i];
}
}
return sum;
}class Solution {
public:
int solution(vector<int> nums) {
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] >= 5) {
sum += nums[i];
}
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] >= 5) {
sum += nums[i];
}
}
return sum;
}
}def solution(nums):
sum = 0
for i in range(len(nums)):
if nums[i] >= 5:
sum += nums[i]
return sumfunction solution(nums) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] >= 5) {
sum += nums[i];
}
}
return sum;
}Asked in Top Tech Interviews
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