Protocol Tome Resolver 49 — Problem Statement & Solution Guide
Problem Description
You are tasked with implementing a resolver function that processes a sequence of integer tokens to compute a specific aggregate metric. Given an array of integers nums and a threshold value K, the resolver must identify all elements in the sequence that are strictly greater than K. The final output is the sum of these qualifying elements. If no elements in the sequence exceed the threshold, the resolver returns 0. This operation simulates a filtering and aggregation step in a data protocol where only high-priority signals (values above K) contribute to the final resolved state.
The input consists of a single array of integers representing the data stream and an integer K representing the activation threshold. The output is a single integer representing the cumulative sum of all values in the array that satisfy the condition value > K. The solution must be efficient, handling large input sizes within standard time complexity limits. Although the problem is categorized under 'Monotonic Stack' patterns in the broader curriculum, this specific instance focuses on the foundational filtering and summation logic that often serves as a base case or preprocessing step in more complex stack-based algorithms.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Tome Resolver 49"
WHY DOES IT MATTER?
Filtering and aggregating data is a foundational pattern for analytics and real‑time monitoring.
OPTIMIZATION CHALLENGE
The key is to avoid extra passes or data structures, reducing the problem to O(n) time and O(1) space.
REAL-WORLD CONNECTION
Think of a firewall that sums the sizes of packets exceeding a bandwidth threshold to trigger alerts.
Initialize the accumulator outside the loop and use a simple if‑condition; avoid premature micro‑optimizations that complicate readability.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a single linear scan where each element is compared against a constant threshold K and, if larger, added to an accumulator. This is a classic example of the "filter‑and‑aggregate" pattern, which can be solved in O(n) time using constant extra space.
Naive alternatives, such as sorting the array first or using nested loops to compare each pair, inflate the time complexity to O(n log n) or O(n²) and are unnecessary because the ordering of elements does not affect the sum of qualifying values. The optimal paradigm leverages the fact that each element is independent, allowing a single pass with a running total.
Interview Questions on This Problem
Q1What is the time and space complexity of summing elements greater than K?
The algorithm runs in O(n) time because it visits each element once. It uses O(1) auxiliary space for the accumulator and loop variables.
Q2Why is sorting the array before summing not beneficial here?
Sorting adds O(n log n) overhead, which is unnecessary since the sum does not depend on order. A linear scan achieves the same result more efficiently.
Q3How would you modify the solution to handle very large sums that might overflow a 32‑bit integer?
Use a 64‑bit integer type (e.g., long long in C++ or long in Java) for the accumulator. Alternatively, employ arbitrary‑precision libraries if the language supports them.
Examples
Input
nums = [12, 5, 8, 20, 3], K = 10
Output
32
Explanation: Iterate through the array: 12 > 10 (add 12), 5 <= 10 (skip), 8 <= 10 (skip), 20 > 10 (add 20), 3 <= 10 (skip). Sum = 12 + 20 = 32.
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 qualify. Sum = 0.
Input
nums = [100, -5, 50, 0, 75], K = 40
Output
175
Explanation: Iterate through the array: 100 > 40 (add 100), -5 <= 40 (skip), 50 > 40 (add 50), 0 <= 40 (skip), 75 > 40 (add 75). Sum = 100 + 50 + 75 = 225. Wait, 100+50+75 is 225. Let me re-calculate. 100+50=150, 150+75=225. Correction: The sum is 225.
Input
nums = [7, 7, 7], K = 6
Output
21
Explanation: Iterate through the array: 7 > 6 (add 7), 7 > 6 (add 7), 7 > 6 (add 7). Sum = 7 + 7 + 7 = 21.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Iterate once, checking each element against K and accumulating the sum directly, achieving O(n) time and O(1) space.
Brute Force Approach
You could sort the array then sum from the first element greater than K, but this adds unnecessary O(n log n) work.
Verified Code Solutions
/**
* @param {number[]} nums
* @param {number} K
* @return {number}
*/
var resolve = function(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
};class Solution {
public:
int resolve(vector<int>& nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
};class Solution {
public int resolve(int[] nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
}class Solution:
def resolve(self, nums: List[int], K: int) -> int:
return sum(num for num in nums if num > K)/**
* @param {number[]} nums
* @param {number} K
* @return {number}
*/
var resolve = function(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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