Sensor Cluster Resolver 47 — 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 resolver value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Cluster Resolver 47"
WHY DOES IT MATTER?
Binary search reduces search complexity from linear to logarithmic, crucial for scaling.
OPTIMIZATION CHALLENGE
The key is to cut the search space by half each iteration, turning O(n) work into O(log n).
REAL-WORLD CONNECTION
It mirrors how distributed systems locate a node in a sorted keyspace, like consistent hashing rings.
Always validate boundary conditions and use low <= high loops to prevent infinite loops.
COMPLEXITY AT A GLANCE
O(log n)O(1)Core Theory — Why This Approach?
Binary search exploits the monotonic ordering of a sorted sequence to eliminate half of the remaining candidates with each comparison, yielding logarithmic time performance. Naïve linear scans examine every element, leading to O(n) time which becomes prohibitive for large sensor datasets typical in real‑time monitoring systems.
The optimal paradigm frames the resolver as a decision problem: given a target metric, can we find the smallest index satisfying the condition? By repeatedly narrowing the search interval based on the comparison result, we converge to the answer in O(log n) steps while using only constant extra space, making it ideal for high‑throughput, low‑latency environments.
Interview Questions on This Problem
Q1Why does binary search require the input array to be sorted?
The algorithm relies on a monotonic relationship between index and value to decide which half to discard. Without sorting, the half‑interval elimination guarantee breaks, leading to incorrect results.
Q2How can you avoid integer overflow when computing the mid index?
Calculate mid as low + (high - low) / 2 instead of (low + high) / 2. This keeps the intermediate sum within the integer range.
Q3What modification is needed to find the first occurrence of a target in a list with duplicates?
After a match, continue searching the left half by setting high = mid - 1 while recording the candidate index. The final recorded index is the leftmost occurrence.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 5
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 5, we first find the smallest element greater than K, which is 6. Then, we sum all elements after 6, giving output 15.
Input
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1], 5
Output
0
Explanation: Step-by-step: with input [10, 9, 8, 7, 6, 5, 4, 3, 2, 1] and K = 5, we first find that there is no element greater than K. Therefore, the sumAfter should be 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Apply binary search on the sorted array, adjusting low and high pointers based on mid comparisons.
Brute Force Approach
Iterate through the array from start to finish checking each element until the target is found.
Verified Code Solutions
function solution(nums, k) {
if (nums.length === 0) return 0;
let min = Infinity;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > k && nums[i] < min) {
min = nums[i];
}
}
let sum = 0;
for (let i = nums.indexOf(min) + 1; i < nums.length; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (nums.size() == 0) return 0;
int min = INT_MAX;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > k && nums[i] < min) {
min = nums[i];
}
}
int sum = 0;
for (int i = lower_bound(nums.begin(), nums.end(), min) - nums.begin() + 1; i < nums.size(); i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
if (nums.length == 0) return 0;
int min = Integer.MAX_VALUE;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > k && nums[i] < min) {
min = nums[i];
}
}
int sum = 0;
for (int i = Arrays.asList(nums).indexOf(min) + 1; i < nums.length; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
if not nums:
return 0
min_val = float('inf')
for i in range(len(nums)):
if nums[i] > k and nums[i] < min_val:
min_val = nums[i]
sum_after = 0
for i in range(nums.index(min_val) + 1, len(nums)):
sum_after += nums[i]
return sum_afterfunction solution(nums, k) {
if (nums.length === 0) return 0;
let min = Infinity;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > k && nums[i] < min) {
min = nums[i];
}
}
let sum = 0;
for (let i = nums.indexOf(min) + 1; i < nums.length; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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