Payload Cipher Resolver 33 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing payload and cipher metrics, and an integer K, construct an optimal algorithm to evaluate and compute the target resolver value under given operational constraints. The algorithm should iterate through the array and add elements that are strictly greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Cipher Resolver 33"
WHY DOES IT MATTER?
Linear‑time aggregation of conditionally filtered data is a foundational pattern for high‑throughput systems.
OPTIMIZATION CHALLENGE
Eliminating extra passes or sorting reduces the algorithm from O(N log N) to O(N).
REAL-WORLD CONNECTION
Network routers sum packet sizes above a threshold to enforce QoS policies.
Keep the loop tight, avoid function calls inside, and use a primitive accumulator to maximize cache friendliness.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The task reduces to a single linear scan where each element is compared against the threshold K and, if greater, contributes to a running total. This paradigm leverages the fact that array elements are independent, allowing O(N) time without auxiliary data structures.
Naïve alternatives—such as nested loops, sorting the array first, or repeatedly scanning for maximums—inflate the time complexity to O(N^2) or O(N log N), which is prohibitive for N up to 10^7. The optimal solution embraces the greedy, one‑pass accumulation model, guaranteeing minimal work per element and constant extra space.
Interview Questions on This Problem
Q1Why is a single pass sufficient to solve this problem?
Each element's contribution depends only on its value relative to K, not on other elements. Therefore, evaluating them independently in one traversal yields the correct sum.
Q2What would be the impact of sorting the array before processing?
Sorting adds O(N log N) overhead, which is unnecessary because order does not affect the sum. The extra time can cause time‑limit failures on large inputs.
Q3How would you handle potential integer overflow in the accumulator?
Use a 64‑bit integer type (e.g., long long in C++ or long in Java) to store the sum. This ensures the accumulator can hold the maximum possible total for the given constraints.
Examples
Input
[30, 40, 50, 20, 10]
Output
120
Explanation: Step-by-step: with input [30, 40, 50, 20, 10], we iterate through the array and add elements that are strictly greater than K (20). So, we add 30, 40, and 50, giving output 120.
Input
[10, 20, 30, 40, 50]
Output
90
Explanation: Step-by-step: with input [10, 20, 30, 40, 50], we iterate through the array and add elements that are strictly greater than K (30). So, we add 40 and 50, giving output 90.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Iterate once, compare each element to K, and add it to a running total if it exceeds K, achieving O(N) time and O(1) extra space.
Brute Force Approach
A brute force method might sort the array then sum, or use nested loops to compare each element multiple times, leading to O(N log N) or O(N^2) 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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