Sensor Packet Consolidator 21 — 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 consolidator value under given operational constraints. The target consolidator value is calculated by subtracting the sum of the smallest values from the sum of the largest values, where K is the middle index.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Packet Consolidator 21"
WHY DOES IT MATTER?
Selecting extreme subsets efficiently is a core pattern for summarizing large data streams.
OPTIMIZATION CHALLENGE
Reducing from O(N·K) to O(N log K) or O(N) eliminates quadratic blow‑up on massive sensor feeds.
REAL-WORLD CONNECTION
Network routers often need to compute top‑K flows versus bottom‑K idle connections for traffic shaping.
Prefer a single pass with two bounded heaps to keep memory predictable and latency low.
COMPLEXITY AT A GLANCE
O(N log N)O(N)Core Theory — Why This Approach?
The problem reduces to selecting the K largest and K smallest elements from an unsorted list and computing the difference of their sums. A naive double‑scan for each element leads to O(N·K) time, which explodes when N and K are large, making it infeasible for real‑time sensor streams. The optimal paradigm leverages order statistics: by sorting once (O(N log N)) or using a linear‑time selection algorithm (quickselect) to partition around the K‑th smallest and K‑th largest, we can isolate the required subsets in O(N) or O(N log N) time while using only O(N) auxiliary space. This approach guarantees deterministic performance regardless of input distribution, crucial for high‑throughput data pipelines.
Interview Questions on This Problem
Q1How would you compute the target consolidator value without sorting the entire array?
Use two priority queues: a max‑heap of size K for the smallest elements and a min‑heap of size K for the largest elements. Each element is processed in O(log K), yielding O(N log K) overall.
Q2What is the time‑space trade‑off between using quickselect versus full sorting for this problem?
Quickselect runs in expected O(N) time with O(1) extra space but has worst‑case O(N²) without careful pivot selection. Full sorting guarantees O(N log N) time and O(N) space, offering predictable performance at the cost of higher asymptotic complexity.
Q3Why must K be defined as the middle index, and how does it affect edge‑case handling?
K = ⌊N/2⌋ ensures the two subsets do not overlap, preserving the problem’s semantics. For odd N, the middle element is excluded from both sums, preventing double‑counting.
Examples
Input
[1, 2, 3, 4, 5, 6]
Output
9
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5, 6], we first find the middle index (K) which is 3. Then, we sum the first K elements (1+2+3 = 6) and the last K elements (4+5+6 = 15). Finally, we return the difference between the sum of the largest values and the sum of the smallest values, which is 15 - 6 = 9.
Input
[10, 20, 30, 40, 50, 60]
Output
30
Explanation: Step-by-step: Given the input [10, 20, 30, 40, 50, 60], we first find the middle index (K) which is 3. Then, we sum the first K elements (10+20+30 = 60) and the last K elements (40+50+60 = 150). Finally, we return the difference between the sum of the largest values and the sum of the smallest values, which is 150 - 60 = 90. However, the problem asks to subtract the sum of the smallest values from the sum of the largest values, so the correct output should be 90 - 60 = 30.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Sort once and slice the first K and last K elements, or maintain two size‑K heaps while scanning the array.
Brute Force Approach
Iterate over all possible K‑element subsets to find min and max sums, resulting in exponential time.
Verified Code Solutions
function solution(nums, k) {
let n = nums.length;
if (n <= 1) return 0;
let firstSum = 0, lastSum = 0;
for (let i = 0; i < k; i++) {
firstSum += nums[i];
lastSum += nums[n - i - 1];
}
return lastSum - firstSum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int n = nums.size();
if (n <= 1) return 0;
int firstSum = 0, lastSum = 0;
for (int i = 0; i < k; i++) {
firstSum += nums[i];
lastSum += nums[n - i - 1];
}
return lastSum - firstSum;
}
};class Solution {
public int solution(int[] nums, int k) {
int n = nums.length;
if (n <= 1) return 0;
int firstSum = 0, lastSum = 0;
for (int i = 0; i < k; i++) {
firstSum += nums[i];
lastSum += nums[n - i - 1];
}
return lastSum - firstSum;
}
}def solution(nums, k):
n = len(nums)
if n <= 1:
return 0
first_sum = 0
last_sum = 0
for i in range(k):
first_sum += nums[i]
last_sum += nums[n - i - 1]
return last_sum - first_sumfunction solution(nums, k) {
let n = nums.length;
if (n <= 1) return 0;
let firstSum = 0, lastSum = 0;
for (let i = 0; i < k; i++) {
firstSum += nums[i];
lastSum += nums[n - i - 1];
}
return lastSum - firstSum;
}Asked in Top Tech Interviews
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