Sensor Cluster Partition 23 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the data aggregation pipeline for a distributed sensor network. The network transmits a stream of integer readings, and a specific filtering threshold K is applied to isolate high-intensity signals. Your objective is to compute the cumulative sum of all readings that strictly exceed the threshold K.
Given an array of integers representing sensor readings and an integer K representing the threshold, return the sum of all elements in the array that are greater than K. If no elements exceed the threshold, return 0.
The solution must efficiently process the array in a single pass, ensuring optimal time complexity for large-scale data streams.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Cluster Partition 23"
WHY DOES IT MATTER?
Trie‑based aggregation turns a linear scan into a constant‑time‑per‑bit operation, dramatically reducing runtime for large numeric datasets.
OPTIMIZATION CHALLENGE
The key is to collapse millions of values into a compact hierarchy where whole sub‑trees can be summed in O(1).
REAL-WORLD CONNECTION
Network routers use prefix trees to route IP packets based on binary prefixes, similar to how we aggregate sensor readings by bit prefixes.
Keep node structures lightweight – store only count and sum – and reuse existing nodes to limit memory overhead.
COMPLEXITY AT A GLANCE
O(N·B) build + O(B) query, where B≈31 for 32‑bit intsO(N·B) nodes (worst case), each storing two integersCore Theory — Why This Approach?
A binary trie (also called a prefix tree) stores each integer as a path of bits from the most‑significant to the least‑significant. By augmenting every node with two aggregates – the number of values that pass through the node and the sum of those values – we can answer range‑sum queries like “sum of all numbers > K” in time proportional to the bit‑length of the integers, independent of the total count of elements.\nNaïve scanning of the entire array for each query is O(N) and becomes prohibitive when N reaches tens of millions or when multiple queries are required. The optimal paradigm leverages the trie’s hierarchical decomposition of the numeric space: during a query we walk the bits of K, and whenever the current bit of K is 0 we can safely add the pre‑computed sum of the entire subtree representing a 1 at that position, thereby skipping large groups of numbers in constant time per bit.
Interview Questions on This Problem
Q1How does augmenting a binary trie with sum information enable O(bit‑length) range‑sum queries?
Each node stores the total of all numbers in its subtree. When traversing the bits of K, we can add the stored sum of a whole subtree whenever the path diverges to a larger value, avoiding individual element visits.
Q2What is the time and space complexity of building and querying such a trie for 32‑bit integers?
Insertion of N numbers costs O(N·32) time and creates at most N·32 nodes. A query also runs in O(32) time, and the extra space is O(N·32) for the nodes plus O(1) per node for aggregates.
Q3Why might a simple sorting‑and‑prefix‑sum solution be less suitable than a trie in a streaming scenario?
Sorting requires all data upfront and O(N log N) time, while a trie can ingest values online in O(1) per bit. It also allows immediate queries without re‑sorting after each insertion.
Examples
Input
readings = [12, 5, 23, 8, 34], K = 10
Output
57
Explanation: Iterate through the array: 12 > 10 (add 12), 5 <= 10 (skip), 23 > 10 (add 23), 8 <= 10 (skip), 34 > 10 (add 34). Sum = 12 + 23 + 34 = 57.
Input
readings = [1, 2, 3, 4, 5], K = 10
Output
0
Explanation: Iterate through the array: All elements (1, 2, 3, 4, 5) are less than or equal to 10. No elements are added to the sum. Final sum = 0.
Input
readings = [100, 200, 300], K = 150
Output
500
Explanation: Iterate through the array: 100 <= 150 (skip), 200 > 150 (add 200), 300 > 150 (add 300). Sum = 200 + 300 = 500.
Input
readings = [-5, -10, 0, 5, 10], K = -1
Output
15
Explanation: Iterate through the array: -5 <= -1 (skip), -10 <= -1 (skip), 0 > -1 (add 0), 5 > -1 (add 5), 10 > -1 (add 10). Sum = 0 + 5 + 10 = 15.
Constraints
- 1 <= readings.length <= 10^5
- -10^9 <= readings[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Build a binary trie with count and sum at each node; query by walking K’s bits and aggregating sums of qualifying sub‑trees – O(bit‑length) per query.
Brute Force Approach
Iterate through the array once, adding each element to the answer if it is greater than K – O(N) time, O(1) extra space.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (num > k) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int> nums, int k) {
int sum = 0;
for (int num : nums) {
if (num > k) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
for (int num : nums) {
if (num > k) {
sum += num;
}
}
return sum;
}
}def solution(nums, k):
sum = 0
for num in nums:
if num > k:
sum += num
return sumfunction solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (num > k) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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