Node Vault Detector 45 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing node and vault metrics, construct an optimal algorithm to evaluate and compute the target detector value under the given operational constraint: sum all elements in the array greater than or equal to K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Vault Detector 45"
WHY DOES IT MATTER?
Binary search turns linear scans into logarithmic look‑ups, essential for high‑throughput query workloads.
OPTIMIZATION CHALLENGE
The key is reducing per‑query work from O(N) to O(log N) via ordering and prefix/suffix aggregates.
REAL-WORLD CONNECTION
Databases use indexed B‑trees to locate rows ≥ a key, mirroring this pattern.
Always sort once, cache cumulative sums, and reuse them across queries to avoid recomputation.
COMPLEXITY AT A GLANCE
O(N log N) preprocessing, O(log N) per queryO(N) for the sorted array and suffix sumsCore Theory — Why This Approach?
The naive solution iterates the array and adds every element that satisfies the condition, which is O(N) but fails when the problem extends to multiple queries or when the array must be pre‑processed for fast repeated look‑ups. By sorting the array once (O(N log N)) we can apply binary search to locate the first index where value ≥ K, turning each query into a logarithmic operation and allowing us to compute the sum of the suffix in O(1) with a pre‑computed suffix‑sum array.
Sorting also imposes an order that enables the binary search paradigm: a divide‑and‑conquer technique that repeatedly halves the search space, guaranteeing worst‑case logarithmic time. This shift from linear scans to logarithmic look‑ups is the cornerstone of many performance‑critical systems where query latency must be minimized.
Interview Questions on This Problem
Q1Why is sorting followed by binary search preferable to scanning the array for each query?
Sorting creates a monotonic order, allowing binary search to locate the threshold in O(log N). This reduces repeated O(N) scans to O(log N) per query, dramatically improving scalability.
Q2How can you compute the sum of all elements ≥ K after finding the start index?
Pre‑compute a suffix‑sum array where suffix[i] = sum of elements from i to end. Once the start index is known, the answer is suffix[start].
Q3What is the overall time complexity for handling Q queries after preprocessing?
Preprocessing takes O(N log N) for sorting plus O(N) for suffix sums. Each query is O(log N), so total is O(N log N + Q log N).
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90]
Output
210
Explanation: Step-by-step: Given the input array [10, 20, 30, 40, 50, 60, 70, 80, 90] and K = 50, we first sort the array in ascending order. Then, we use binary search to find the index of the first element greater than or equal to K. Finally, we sum all elements in the array from the found index to the end.
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5] and K = 3, we first sort the array in ascending order. Then, we use binary search to find the index of the first element greater than or equal to K. Finally, we sum all elements in the array from the found index to the end.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Sort the array, compute suffix sums, then binary‑search the lower bound for K and return the pre‑computed suffix sum.
Brute Force Approach
Iterate through the array and sum every element that is ≥ K, which is O(N) per query.
Verified Code Solutions
function solution(nums, K) {
nums.sort((a, b) => a - b);
let left = 0, right = nums.length - 1;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] < K) left = mid + 1;
else right = mid - 1;
}
let sum = 0;
for (let i = left; i < nums.length; i++) sum += nums[i];
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
sort(nums.begin(), nums.end());
int left = 0, right = nums.size() - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] < K) left = mid + 1;
else right = mid - 1;
}
int sum = 0;
for (int i = left; i < nums.size(); i++) sum += nums[i];
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
Arrays.sort(nums);
int left = 0, right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] < K) left = mid + 1;
else right = mid - 1;
}
int sum = 0;
for (int i = left; i < nums.length; i++) sum += nums[i];
return sum;
}
}def solution(nums, K):
nums.sort()
left, right = 0, len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] < K: left = mid + 1
else: right = mid - 1
sum = 0
for i in range(left, len(nums)):
sum += nums[i]
return sumfunction solution(nums, K) {
nums.sort((a, b) => a - b);
let left = 0, right = nums.length - 1;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] < K) left = mid + 1;
else right = mid - 1;
}
let sum = 0;
for (let i = left; i < nums.length; i++) sum += nums[i];
return sum;
}Asked in Top Tech Interviews
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