Node Vault Aligner 2 — Problem Statement & Solution Guide
Problem Description
In a distributed storage network, a sequence of integer metrics represents the capacity load on individual nodes. An alignment threshold K is defined to identify overloaded segments. Your task is to process the sequence and compute the aggregate capacity of all nodes that strictly exceed this threshold. Specifically, given an array of integers and a target value K, return the sum of all elements in the array that are greater than K. If no elements exceed K, return 0.
The input consists of a single array of integers representing the node metrics and an integer K representing the alignment threshold. The output is a single integer representing the sum of qualifying metrics. This problem requires a linear scan of the data to identify and accumulate the relevant values efficiently.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Vault Aligner 2"
WHY DOES IT MATTER?
Linear scans are the most efficient pattern for threshold‑based aggregation because they avoid unnecessary sorting or data structure overhead, ensuring O(n) time and O(1) space. This pattern is essential in real‑time monitoring systems where latency and memory footprint are critical.
OPTIMIZATION CHALLENGE
The key insight is that you only need to compare each element once; no sorting or additional data structures are required. This reduces both time and space complexity dramatically.
REAL-WORLD CONNECTION
In distributed storage, monitoring node loads often requires summing capacities that exceed a safety threshold. A single pass over the metrics stream allows operators to trigger alerts immediately without storing the entire dataset.
When implementing, use a simple for‑loop with an early exit for empty arrays, and consider using 64‑bit integers to avoid overflow when summing large values.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to computing the sum of all array elements that exceed a given threshold K. A naive approach might involve nested loops or repeated scans, leading to O(n^2) time if one mistakenly recomputes partial sums for each element. The optimal strategy is a single linear pass: iterate once, add any element greater than K to an accumulator. If the input array is guaranteed sorted, we can further optimize by performing a binary search to locate the first element > K and then summing the suffix, which still requires O(n) for the sum but reduces the number of comparisons to O(log n). This linear-time, constant-space solution is the standard paradigm for threshold‑based aggregation problems.
Interview Questions on This Problem
Q1How would you handle large datasets where the array cannot fit into memory?
Use a streaming approach: read the data in chunks, maintain a running sum of elements greater than K, and discard each chunk after processing. This keeps memory usage constant regardless of input size.
Q2If the array is sorted in ascending order, how can you reduce the number of elements you need to examine?
Perform a binary search to find the first element strictly greater than K. Once located, sum the remaining suffix of the array. This reduces the number of comparisons to O(log n) before the linear sum of the suffix.
Q3What if the threshold K changes frequently? How would you design a system to answer queries quickly?
Precompute a prefix sum array or a balanced BST keyed by element values. For each query, use binary search on the sorted array to find the split point and then compute the sum using the prefix sums, giving O(log n) query time.
Examples
Input
nums = [12, 5, 8, 15, 3], K = 10
Output
27
Explanation: Iterate through the array: 12 > 10 (add 12), 5 <= 10 (skip), 8 <= 10 (skip), 15 > 10 (add 15), 3 <= 10 (skip). Sum = 12 + 15 = 27.
Input
nums = [1, 2, 3, 4, 5], K = 10
Output
0
Explanation: Iterate through the array: All elements (1, 2, 3, 4, 5) are less than or equal to 10. No elements are added to the sum. Sum = 0.
Input
nums = [100, 200, 300], K = 50
Output
600
Explanation: Iterate through the array: 100 > 50 (add 100), 200 > 50 (add 200), 300 > 50 (add 300). Sum = 100 + 200 + 300 = 600.
Input
nums = [-5, -1, 0, 5, 10], K = 0
Output
15
Explanation: Iterate through the array: -5 <= 0 (skip), -1 <= 0 (skip), 0 <= 0 (skip), 5 > 0 (add 5), 10 > 0 (add 10). Sum = 5 + 10 = 15.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Traverse the array once, adding elements that are greater than K to a running total, achieving O(n) time and O(1) space.
Brute Force Approach
Loop over each element, and for each element, loop over the entire array again to recompute the sum of elements greater than K, resulting in O(n^2) time.
Verified Code Solutions
function solution(nums, k) { return nums.filter(num => num > k).reduce((a, b) => a + b, 0); }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): return sum(num for num in nums if num > k)function solution(nums, k) { return nums.filter(num => num > k).reduce((a, b) => a + b, 0); }Asked in Top Tech Interviews
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