Payload Cipher Extractor 38 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing payload and cipher metrics, construct an optimal algorithm to evaluate and compute the target extractor value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Cipher Extractor 38"
WHY DOES IT MATTER?
Maximum closure captures many real‑world resource‑selection problems in a single graph formulation.
OPTIMIZATION CHALLENGE
Transforming the combinatorial selection into a min‑cut shrinks exponential search to polynomial max‑flow.
REAL-WORLD CONNECTION
It mirrors selecting secure modules in a software system where each module has benefit and risk, and dependencies must be honored.
Reuse a well‑tested max‑flow library and pre‑compress parallel edges to avoid unnecessary overhead.
COMPLEXITY AT A GLANCE
O(V^2 * E) or O(E * sqrt(V)) with Dinic/Push‑RelabelO(V + E)Core Theory — Why This Approach?
The problem can be modeled as a maximum weight closure in a directed graph, where each vertex carries a payload (positive weight) and a cipher cost (negative weight). By constructing a flow network with a source connected to all positive‑weight vertices and a sink connected from all negative‑weight vertices, the minimum s‑t cut corresponds to the optimal subset of vertices that maximizes total payload minus total cipher cost while respecting the directed dependencies. Naïve enumeration of all vertex subsets is O(2^N) and quickly becomes infeasible for large graphs, especially when the graph contains cycles or dense connectivity. The optimal paradigm leverages the max‑flow/min‑cut theorem, reducing the exponential combinatorial search to a polynomial‑time max‑flow computation, typically using Dinic’s or Push‑Relabel algorithms, which efficiently handle the required capacity constraints and preserve the closure property.
Interview Questions on This Problem
Q1What is the maximum weight closure problem and how does it relate to min‑cut?
It asks for a vertex subset closed under outgoing edges that maximizes the sum of vertex weights. By adding source edges for positive weights and sink edges for negative weights, the min‑cut of the resulting flow network yields the optimal closure.
Q2Why must we connect positive‑weight vertices to the source and negative‑weight vertices to the sink?
Connecting positives to the source forces the cut to pay a capacity equal to the lost payload if the vertex is excluded, while connecting negatives to the sink charges the cut for including costly vertices. This encoding ensures the cut cost equals total penalty of the chosen subset.
Q3How does the presence of cycles affect the closure property and the flow construction?
Cycles do not break closure because any vertex reachable from a selected vertex must also be selected; the flow network respects this via directed edges with infinite capacity. Infinite capacities prevent the cut from separating vertices that would violate the closure constraint.
Examples
Input
[10, 20, 30, 40, 50], 50
Output
90
Explanation: Step-by-step: Given the input array [10, 20, 30, 40, 50] and K = 50, we iterate through the array. We add 10, 20, 30, and 40 to the sum because they are less than K. We do not add 50 to the sum because it is equal to K. Therefore, the output is 90.
Input
[-10, -20, -30, -40], 0
Output
-10
Explanation: Step-by-step: Given the input array [-10, -20, -30, -40] and K = 0, we iterate through the array. We add -10 to the sum because it is less than K. Therefore, the output is -10.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Build a flow network with source/sink edges for weights and infinite edges for dependencies, then run a max‑flow algorithm and take the source side of the min‑cut.
Brute Force Approach
Enumerate every subset of vertices, check closure, and compute net value; exponential 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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