Node Vault Tracker 43 — 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 tracker value under given operational constraints. The target tracker value is the count of sub-arrays with at least one element greater than a given threshold K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Vault Tracker 43"
WHY DOES IT MATTER?
Transforming a counting problem into its complement often reduces quadratic work to linear.
OPTIMIZATION CHALLENGE
The key is to replace nested loops with a single pass that aggregates segment lengths.
REAL-WORLD CONNECTION
Similar to monitoring server logs: instead of scanning every request for an error, count error‑free intervals and infer error occurrences.
Maintain a running length counter; reset it when you encounter a value > K to avoid extra state.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The naive solution enumerates every possible sub‑array, checks its maximum, and increments a counter when the maximum exceeds K. This O(N^2) approach quickly becomes infeasible for N up to 10^5 or higher because the number of sub‑arrays grows quadratically. The optimal paradigm leverages the complement: instead of counting sub‑arrays that contain a value > K, count those that contain only values ≤ K and subtract from the total number of sub‑arrays N·(N+1)/2. By scanning the array once and aggregating lengths of contiguous segments where all elements are ≤ K, we can compute the complement in O(N) time and O(1) extra space, which scales to massive inputs.
Interview Questions on This Problem
Q1How can you count sub‑arrays with at least one element greater than K in linear time?
Compute total sub‑arrays N·(N+1)/2, then subtract the count of sub‑arrays where every element ≤ K. The latter is obtained by summing len·(len+1)/2 for each maximal segment of ≤ K elements.
Q2Why does the complement method avoid double‑counting?
Every sub‑array either has a max > K or all elements ≤ K, never both, so the two sets partition the whole space. Subtracting the size of one partition from the total yields the exact size of the other.
Q3What edge case must you handle when the array contains no element > K?
All elements belong to ≤ K segments, so the complement count equals the total sub‑arrays and the answer becomes zero. Ensure the algorithm correctly returns zero rather than a negative value.
Examples
Input
[1, 2, 3, 4, 5] and K = 5
Output
2
Explanation: Step-by-step: We iterate through the array and count the number of sub-arrays with at least one element greater than K. For the given input, the sub-arrays [3, 4, 5] and [4, 5] satisfy this condition, so the output is 2.
Input
[1, 2, 3, 4, 5] and K = 3
Output
5
Explanation: Step-by-step: We iterate through the array and count the number of sub-arrays with at least one element greater than K. For the given input, all sub-arrays satisfy this condition, so the output is 5.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Traverse once, track lengths of contiguous ≤ K segments, sum their sub‑array counts, and subtract from total sub‑arrays.
Brute Force Approach
Generate every sub‑array, compute its maximum, and increment the answer if the maximum > K.
Verified Code Solutions
function solution(nums, K) {
let count = 0;
for (let i = 0; i < nums.length; i++) {
for (let j = i + 1; j <= nums.length; j++) {
let subArray = nums.slice(i, j);
if (subArray.some(x => x > K)) {
count++;
}
}
}
return count;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int count = 0;
for (int i = 0; i < nums.size(); i++) {
for (int j = i + 1; j <= nums.size(); j++) {
vector<int> subArray(nums.begin() + i, nums.begin() + j);
if (any_of(subArray.begin(), subArray.end(), [K](int x) { return x > K; })) {
count++;
}
}
}
return count;
}
};class Solution {
public int solution(int[] nums, int K) {
int count = 0;
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j <= nums.length; j++) {
int[] subArray = Arrays.copyOfRange(nums, i, j);
if (Arrays.stream(subArray).anyMatch(x -> x > K)) {
count++;
}
}
}
return count;
}
}def solution(nums, K):
count = 0
for i in range(len(nums)):
for j in range(i + 1, len(nums) + 1):
subArray = nums[i:j]
if any(x > K for x in subArray):
count += 1
return countfunction solution(nums, K) {
let count = 0;
for (let i = 0; i < nums.length; i++) {
for (let j = i + 1; j <= nums.length; j++) {
let subArray = nums.slice(i, j);
if (subArray.some(x => x > K)) {
count++;
}
}
}
return count;
}Asked in Top Tech Interviews
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