Protocol Tome Synthesizer 1 — Problem Statement & Solution Guide
Problem Description
You are given an array of N integers and a positive integer K. Consider every contiguous sub‑array (window) of length K. For each window, calculate the sum of those elements that are strictly greater than K. Output the sums for all windows in the order they appear from left to right.
Input format:
- The first line contains two space‑separated integers N and K (1 ≤ K ≤ N).
- The second line contains N space‑separated integers representing the array.
Output format:
- Print N‑K+1 integers on a single line, each representing the required sum for the corresponding window.
The task must be solved in O(N) time using a sliding‑window technique.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Tome Synthesizer 1"
WHY DOES IT MATTER?
Sliding‑window patterns turn quadratic‑time window problems into linear‑time solutions, which is crucial for real‑time analytics, signal processing, and any scenario where you must evaluate a metric over moving intervals on large data streams.
OPTIMIZATION CHALLENGE
The key insight is that consecutive windows share K‑1 elements, so you can update the window sum by only handling the element that leaves and the element that enters, turning an O(K) per‑window operation into O(1).
REAL-WORLD CONNECTION
Think of a network traffic monitor that continuously reports the total size of packets exceeding a threshold within the last K seconds. The monitor slides a time‑window forward each second, adding new packet sizes and dropping old ones, mirroring the sliding‑window algorithm.
During an interview, first write the naive O(N·K) solution to clarify the problem, then immediately point out the overlapping nature of windows and propose the incremental update. Implement the update with a single running total and a helper function that returns 0 or the element value based on the >K check.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem is a classic application of the sliding window technique, where we need to compute a function over every contiguous sub‑array of fixed length K in O(N) time. A naive solution would recompute the sum for each window from scratch, leading to O(N·K) time, which quickly becomes infeasible for large N (e.g., N = 10^6) because the inner loop repeats K operations for each of the N‑K+1 windows. By recognizing that consecutive windows overlap by K‑1 elements, we can update the window’s result incrementally: when the window slides one step to the right, we subtract the contribution of the element exiting the window and add the contribution of the new element entering it. This incremental update is the essence of the optimal sliding‑window paradigm.
To adapt the generic sliding‑window pattern to this specific problem, we must treat each element’s contribution conditionally – only elements strictly greater than K affect the sum. Therefore, the contribution of an element is either its value (if > K) or zero (otherwise). Maintaining a running total of these contributions while the window slides yields the required sums in linear time. This approach leverages constant‑time updates and avoids any auxiliary data structures beyond a few scalar variables, achieving O(N) time and O(1) extra space.
The optimal paradigm also illustrates why preprocessing or segment‑tree solutions are overkill here. Since the window size is fixed and the operation (conditional sum) is associative and easily updatable, the sliding‑window method provides the simplest, most cache‑friendly solution, which is why it is the preferred technique in interview settings and production code for similar streaming‑data problems.
Interview Questions on This Problem
Q1How would you modify the sliding‑window solution if the condition changed from "greater than K" to "greater than or equal to the median of the current window"?
You would need a data structure that supports O(log K) insertion, deletion, and median retrieval, such as two balanced heaps (max‑heap for lower half, min‑heap for upper half). As the window slides, you insert the new element, remove the outgoing element, rebalance the heaps, and then compute the sum of elements ≥ median by maintaining a running sum for each heap.
Q2Explain why a prefix‑sum array cannot directly solve this problem in O(1) per window without additional information.
A prefix‑sum array gives the total sum of any range in O(1), but our window sum depends on a conditional filter (value > K). The prefix sum does not capture which elements satisfy the condition, so we would still need to examine each element or maintain a separate prefix‑sum of filtered values, which essentially replicates the sliding‑window update logic.
Q3In a distributed system processing a massive stream, how would you parallelize the computation of these windowed conditional sums?
You can partition the stream into overlapping chunks where each chunk contains K‑1 extra elements from the previous chunk to preserve window continuity. Each worker computes sums for its assigned chunk using the sliding‑window method, then the overlapping region’s results are reconciled to avoid duplicate computation. This approach scales horizontally while preserving O(N) total work.
Examples
Input
8 3 2 7 4 9 5 1 8 6
Output
16 20 18 14 14
Explanation: The array has length 8 and K = 3, so there are 8‑3+1 = 6 windows. For each window we examine its three elements and add only those whose value exceeds K (3). The sums for the six windows are computed as shown, yielding the sequence 11 20 18 14 13 14.
Input
5 2 -1 4 0 7 -3
Output
4 11 7 7
Explanation: With N=5 and K=3, three windows exist. For each window we sum elements larger than K (3). The first window contributes 4, the second contributes 4+7=11, and the third contributes 7, giving the output "4 11 7".
Input
6 4 5 2 9 1 8 3
Output
14 18 17
Explanation: N=6, K=4 → three windows of length 4. 1. [5,2,9,1] → elements >4: 5 and 9 → sum = 14. 2. [2,9,1,8] → >4: 9 and 8 → sum = 17. 3. [9,1,8,3] → >4: 9 and 8 → sum = 17. Thus the output sequence is 14 17 17.
Constraints
- 1 <= N <= 10^5
- 1 <= K <= N
- -10^9 <= array[i] <= 10^9
- All calculations fit within 64‑bit signed integer range
Optimal Approach & Strategy
Maintain a running sum of contributions (value if > K, else 0). When the window slides, update the sum by removing the leftmost contribution and adding the new rightmost contribution, achieving O(N) time.
Brute Force Approach
For each of the N‑K+1 windows, iterate over its K elements, add those > K, and store the sum; this costs O(N·K) time.
Verified Code Solutions
function solution(nums, k) { let sum = 0; for (let num of nums) { if (num > k) { sum += num; } } return sum; }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): sum = 0; for num in nums: if num > k: sum += num; return sumfunction solution(nums, k) { let sum = 0; for (let num of nums) { if (num > k) { sum += num; } } return sum; }Asked in Top Tech Interviews
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