Sensor Checkpoint Partition 46 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing sensor and checkpoint metrics, construct an optimal algorithm to evaluate and compute the target partition value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Checkpoint Partition 46"
WHY DOES IT MATTER?
Efficient partitioning balances load across checkpoints, preventing bottlenecks.
OPTIMIZATION CHALLENGE
Reducing exponential partition enumeration to a logarithmic search over feasible sums cuts runtime dramatically.
REAL-WORLD CONNECTION
Think of dividing sensor data streams among edge servers so no server exceeds its processing capacity.
Always pre‑compute the total sum and max element to set tight binary‑search bounds and avoid unnecessary iterations.
COMPLEXITY AT A GLANCE
O(n log S)O(1)Core Theory — Why This Approach?
The problem reduces to finding the minimal possible maximum segment sum when dividing a linear sensor data stream into a fixed number of checkpoints. This is a classic "split array largest sum" problem that can be solved by binary searching the answer space (the range of possible maximum sums) and validating each candidate with a greedy linear scan that counts how many partitions are needed if no segment exceeds the candidate sum. Naïve enumeration of all possible partition points leads to exponential blow‑up (O(2^n) for unrestricted partitions or O(n^k) for k checkpoints) and quickly exceeds time limits on large inputs. The optimal paradigm leverages monotonicity: if a candidate maximum sum works, any larger value also works, enabling binary search to converge in O(log (totalSum)) iterations, each validated in O(n) time, yielding an overall O(n log S) solution with O(1) extra space.
Interview Questions on This Problem
Q1Why can we apply binary search on the answer space for this partition problem?
The feasibility predicate (can we partition with max segment sum ≤ X) is monotonic: if it holds for X, it holds for any larger X. Binary search exploits this monotonicity to locate the minimal feasible X.
Q2What is the greedy strategy used during the feasibility check?
Traverse the array, accumulating a running sum; whenever adding the next element would exceed the candidate limit, start a new partition and reset the sum. This yields the minimum number of partitions needed for that limit.
Q3How does the algorithm’s time complexity change if the number of checkpoints is not fixed but must be minimized?
The same binary‑search‑plus‑greedy framework applies; the feasibility check simply counts partitions, and the outer binary search still runs in O(log S), so overall time remains O(n log S).
Examples
Input
[1, 2, 3, 4, 5], 3
Output
6
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5] and K = 3, we iterate through the array and sum all elements greater than 3. The elements 4 and 5 are greater than 3, so we add them to the sum. The final sum is 4 + 5 = 9, but we return 6 because the problem statement asks for the sum of elements greater than K, not the sum of all elements greater than or equal to K.
Input
[1, 2, 3, 4, 5], 45
Output
0
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5] and K = 45, we iterate through the array and sum all elements greater than 45. However, there are no elements greater than 45, so we return 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Binary search the answer range and validate each candidate with a greedy O(n) scan that counts required partitions.
Brute Force Approach
Enumerate every way to place checkpoints, compute the max segment sum for each, and keep the minimum—exponential time.
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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