Pipeline Beacon Tracker 32 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and beacon metrics, construct an optimal algorithm to evaluate and compute the target tracker value under given operational constraints. The input array should be sorted in ascending order. The tracker value should be initialized to 0. Iterate through the sorted array and add each number greater than K to the tracker value.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Beacon Tracker 32"
WHY DOES IT MATTER?
This pattern tests the candidate's ability to recognize when complex data structures (like Tries or Segment Trees) are overkill for a problem that can be solved with basic array properties (sorted order) and linear or logarithmic scans. It highlights the importance of reading constraints carefully.
OPTIMIZATION CHALLENGE
The key insight is that the array is sorted. This allows the use of binary search to find the partition point where elements become greater than K, reducing the search space from O(N) to O(log N) for locating the start of the valid range. If prefix sums are precomputed, the summation itself becomes O(1).
REAL-WORLD CONNECTION
In distributed systems, logs are often sorted by timestamp. When filtering for 'critical' events (value > K), engineers often use binary search on sorted indices to quickly jump to the relevant section of the log, rather than scanning from the beginning, especially when the log is stored in a columnar database or sorted file.
In an interview, do not jump to building a Trie or a complex tree. Explicitly state: 'Since the array is sorted, I can use binary search to find the first element greater than K. If I need to sum the rest, I can either iterate from that index or use a precomputed suffix sum array for O(1) retrieval.' This shows awareness of both simple and optimized paths.
COMPLEXITY AT A GLANCE
O(N) for single query without prefix sums, O(log N) for query with precomputed prefix sumsO(1) for single query, O(N) for precomputed prefix sumsCore Theory — Why This Approach?
The problem 'Pipeline Beacon Tracker 32' presents a deceptively simple constraint: summing elements greater than a threshold K from a sorted array. While the core operation is linear, the theoretical foundation lies in understanding the properties of sorted data structures and the efficiency of linear scans versus more complex data structures like Tries. In a standard unsorted scenario, one might consider building a Trie or a Binary Search Tree to facilitate range queries or prefix sums, but the explicit constraint that the input is already sorted in ascending order renders such complex structures unnecessary for this specific query type. The optimal paradigm here is the direct linear scan, leveraging the sorted nature to potentially allow for early termination if the array were descending, or simply to guarantee that once an element exceeds K, all subsequent elements also exceed K (if we were looking for a contiguous suffix), though in this specific 'sum all > K' case, we must still verify each element unless we use binary search to find the first index > K and then sum the suffix.
Interview Questions on This Problem
Q1At a fintech platform, you need to calculate the total value of transactions exceeding a fraud detection threshold K from a log that is already sorted by timestamp (and thus value in this specific metric). How would you optimize this if the log is massive and you need to perform this query repeatedly with different K values?
For a single query on a sorted array, a linear scan is O(N). However, if queries are repeated, one could precompute a prefix sum array. Then, for a given K, use binary search to find the first index where value > K, and subtract the prefix sum at that index from the total sum. This reduces each query to O(log N) time after O(N) preprocessing.
Q2In a high-growth startup, you are building a real-time dashboard that displays the sum of active pipeline deals greater than a dynamic threshold. The data stream is sorted by deal size. How do you handle the 'greater than K' condition efficiently without re-scanning the entire list every time K changes slightly?
Since the data is sorted, you can maintain a pointer to the first element greater than the current K. If K increases, move the pointer forward until the condition is met. If K decreases, you might need to move the pointer backward or re-evaluate. For a static sorted array, binary search is the most robust approach to locate the boundary index in O(log N), followed by a suffix sum calculation if precomputed, or a linear sum of the suffix if not.
Q3A global product company asks you to implement a function that returns the sum of all numbers in a sorted array that are strictly greater than K. What is the time complexity of your solution, and can you improve it if the array is static and queries are frequent?
The basic solution is O(N) using a linear scan. If the array is static and queries are frequent, precompute a prefix sum array in O(N) space. Then, for each query, use binary search to find the lower bound of K (first element > K) in O(log N), and compute the sum as total_sum - prefix_sum[index]. This optimizes the query time to O(log N).
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
Output
100
Explanation: Step 1: Sort the input array in ascending order. Input: [10, 20, 30, 40, 50, 60, 70, 80, 90, 100]. Output: [10, 20, 30, 40, 50, 60, 70, 80, 90, 100]. Step 2: Initialize the tracker value to 0. Tracker value: 0. Step 3: Iterate through the sorted array and add each number greater than K to the tracker value. For this example, let's assume K = 50. Tracker value: 50 (10 + 20 + 30 + 40) + 60 (greater than K) + 70 (greater than K) + 80 (greater than K) + 90 (greater than K) + 100 (greater than K). Final tracker value: 500.
Input
[5, 15, 25, 35, 45, 55, 65, 75, 85, 95]
Output
45
Explanation: Step 1: Sort the input array in ascending order. Input: [5, 15, 25, 35, 45, 55, 65, 75, 85, 95]. Output: [5, 15, 25, 35, 45, 55, 65, 75, 85, 95]. Step 2: Initialize the tracker value to 0. Tracker value: 0. Step 3: Iterate through the sorted array and add each number greater than K to the tracker value. For this example, let's assume K = 50. Tracker value: 45 (5 + 15 + 25 + 35). Final tracker value: 45.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use binary search to find the first index where the element is greater than K. Then, sum the elements from that index to the end of the array. If prefix sums are precomputed, the sum can be retrieved in constant time.
Brute Force Approach
Iterate through the entire array from start to finish, checking each element to see if it is greater than K. If it is, add it to a running total; otherwise, skip it.
Verified Code Solutions
function solution(nums, K) {
if (nums.length === 0) return 0;
nums.sort((a, b) => a - b);
let tracker = 0;
for (let num of nums) {
if (num > K) tracker += num;
}
return tracker;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
if (nums.size() == 0) return 0;
sort(nums.begin(), nums.end());
int tracker = 0;
for (int num : nums) {
if (num > K) tracker += num;
}
return tracker;
}
};class Solution {
public int solution(int[] nums, int K) {
if (nums.length == 0) return 0;
Arrays.sort(nums);
int tracker = 0;
for (int num : nums) {
if (num > K) tracker += num;
}
return tracker;
}
}def solution(nums, K):
if not nums:
return 0
nums.sort()
tracker = 0
for num in nums:
if num > K:
tracker += num
return trackerfunction solution(nums, K) {
if (nums.length === 0) return 0;
nums.sort((a, b) => a - b);
let tracker = 0;
for (let num of nums) {
if (num > K) tracker += num;
}
return tracker;
}Asked in Top Tech Interviews
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