Sensor Cluster Consolidator 9 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing sensor and cluster metrics, construct an optimal algorithm to evaluate and compute the target consolidator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Cluster Consolidator 9"
WHY DOES IT MATTER?
Tree DP transforms exponential sub‑problem overlap into linear work.
OPTIMIZATION CHALLENGE
The key is to propagate only the minimal necessary state up the tree to cut the complexity from quadratic to linear.
REAL-WORLD CONNECTION
It mirrors aggregating metrics in hierarchical IoT gateways before sending to the cloud.
Cache child results in a struct and avoid mutable global state to keep the recursion clean and debuggable.
COMPLEXITY AT A GLANCE
O(N)O(H)Core Theory — Why This Approach?
The problem maps naturally to a binary tree where each sensor metric becomes a node and cluster relationships define parent‑child links. A bottom‑up DP using post‑order traversal aggregates child contributions to compute the consolidator value in linear time, exploiting the tree’s hierarchical substructure. Naïve approaches—such as recomputing sub‑cluster values for every node or using repeated scans of the input—inflate the runtime to O(N²) because each node’s subtree is visited many times, which quickly exceeds limits on large sensor networks. The optimal paradigm treats the tree as a directed acyclic graph, processes each node exactly once, and carries forward only the necessary aggregates, achieving O(N) time and O(H) auxiliary space where H is the tree height.
Interview Questions on This Problem
Q1How does post‑order traversal enable O(N) computation of the consolidator value?
Post‑order visits children before their parent, ensuring all sub‑cluster aggregates are ready when the parent is processed. This eliminates redundant recomputation and guarantees each node contributes exactly once.
Q2What space complexity does a recursive solution incur, and how can it be reduced?
A recursive solution uses O(H) call‑stack space, where H is the tree height. Converting to an explicit stack or Morris traversal can bring auxiliary space down to O(1) while preserving linear time.
Q3Why does a naïve double‑loop over the input sequence fail for large N?
The double‑loop recomputes overlapping sub‑problems, leading to O(N²) operations that exceed time limits for big sensor arrays. Dynamic programming or tree DP collapses these overlaps into a single linear pass.
Examples
Input
[50, 50, 40, 40, 30, 20, 10, 5, 2, 1], 3
Output
150
Explanation: Step-by-step: First, sort the input array in ascending order. Then, calculate the sum of the first K elements in the sorted array. In this case, the sum is 50 + 50 + 40 = 140. However, since K=3 is not enough to cover all the elements in the array, we need to include the next element 40 in the sum. Therefore, the final output is 140 + 40 = 150.
Input
[50, 40, 40, 30, 20, 10, 5, 2, 1], 3
Output
130
Explanation: Step-by-step: First, sort the input array in ascending order. Then, calculate the sum of the first K elements in the sorted array. In this case, the sum is 50 + 40 + 40 = 130, which is correct.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Construct the tree and perform a single post‑order traversal that aggregates child results, achieving O(N) time.
Brute Force Approach
Repeatedly scan the list to recompute each cluster’s value from scratch, leading to O(N²) time.
Verified Code Solutions
function solution(nums, k) {
if (k >= nums.length) return nums.reduce((a, b) => a + b, 0);
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
if (k < nums.length) {
sum += nums[k - 1];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (k >= nums.size()) return accumulate(nums.begin(), nums.end(), 0);
sort(nums.begin(), nums.end());
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
if (k < nums.size()) {
sum += nums[k - 1];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
if (k >= nums.length) return Arrays.stream(nums).sum();
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
if (k < nums.length) {
sum += nums[k - 1];
}
return sum;
}
}def solution(nums, k):
if k >= len(nums):
return sum(nums)
nums.sort()
total = 0
for i in range(k):
total += nums[i]
if k < len(nums):
total += nums[k - 1]
return totalfunction solution(nums, k) {
if (k >= nums.length) return nums.reduce((a, b) => a + b, 0);
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
if (k < nums.length) {
sum += nums[k - 1];
}
return sum;
}Asked in Top Tech Interviews
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