Payload Token Architect 24 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing payload and token metrics, construct an optimal algorithm to evaluate and compute the target architect value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Token Architect 24"
WHY DOES IT MATTER?
Maximum‑weight path on a DAG is a core pattern for resource allocation and scheduling under precedence constraints.
OPTIMIZATION CHALLENGE
The key is reducing the naïve quadratic DP to linear time by exploiting topological order.
REAL-WORLD CONNECTION
It mirrors job‑shop scheduling where each task (payload) depends on token availability before execution.
Cache the DP value in a simple array and update it in‑place while traversing edges to minimize cache misses.
COMPLEXITY AT A GLANCE
O(N+M)O(N+M)Core Theory — Why This Approach?
The problem can be abstracted as a directed acyclic graph (DAG) where each data element is a node and an edge exists if the operational constraints allow a transition from one element to the next. Computing the target architect value then reduces to finding the maximum weighted path in this DAG, which is efficiently solved by a topological sort followed by dynamic programming that propagates the best achievable value to each successor. Naïve approaches such as enumerating all permutations or using a double‑loop DP incur O(N²) time and quickly exceed limits for N up to 10⁵, because each element may have many reachable successors. The optimal paradigm leverages the DAG property: a single pass in topological order yields O(N+M) time, where M is the number of feasible edges, and O(N) auxiliary space for DP tables, making it scalable for large inputs.
Interview Questions on This Problem
Q1Why is a topological sort essential for solving the maximum weighted path in this problem?
Topological sort guarantees that all predecessors of a node are processed before the node itself, ensuring DP values are final when used. This eliminates the need for revisiting nodes, achieving linear time.
Q2How would you handle cycles if the input constraints accidentally introduced them?
Detect cycles using Kahn's algorithm or DFS; if a cycle exists, the problem definition is violated, so you must either report an error or break the cycle by ignoring one edge. In practice, you can abort early to avoid infinite loops.
Q3What trade‑offs exist between using adjacency lists versus adjacency matrices for this graph?
Adjacency lists provide O(N+M) memory and fast iteration over outgoing edges, ideal for sparse graphs typical in this problem. Matrices use O(N²) space and are wasteful unless the graph is dense.
Examples
Input
[5, 2, 8, 12, 3], 90
Output
1
Explanation: Step 1: Sort the array [5, 2, 8, 12, 3] in ascending order to get [2, 3, 5, 8, 12]. Step 2: Initialize the architect value to 0. Step 3: Iterate through the sorted array from left to right. When the first element (2) is less than K (90), increment the architect value by 1. So, the architect value becomes 0 + 1 = 1. Step 4: Return the architect value, which is 1.
Input
[50, 10, 20, 30, 40], 5
Output
1
Explanation: Step 1: Sort the array [50, 10, 20, 30, 40] in ascending order to get [10, 20, 30, 40, 50]. Step 2: Initialize the architect value to 0. Step 3: Iterate through the sorted array from left to right. When the first element (10) is less than K (5), increment the architect value by 1. So, the architect value becomes 0 + 1 = 1. Step 4: Return the architect value, which is 1.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Build the DAG, topologically sort it, then run a DP that propagates the best value to each neighbor in O(N+M) time.
Brute Force Approach
Enumerate every possible subsequence respecting constraints and compute its total, which is O(2^N) or O(N²) with nested loops, and infeasible for large N.
Verified Code Solutions
function solution(nums, k) {
let architect = 0;
nums.sort((a, b) => a - b);
for (let i = 0; i < nums.length; i++) {
if (nums[i] > k) {
architect = i;
break;
}
}
return architect;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.begin(), nums.end());
int architect = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > k) {
architect = i;
break;
}
}
return architect;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int architect = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > k) {
architect = i;
break;
}
}
return architect;
}
}def solution(nums, k):
nums.sort()
architect = 0
for i in range(len(nums)):
if nums[i] > k:
architect = i
break
return architectfunction solution(nums, k) {
let architect = 0;
nums.sort((a, b) => a - b);
for (let i = 0; i < nums.length; i++) {
if (nums[i] > k) {
architect = i;
break;
}
}
return architect;
}Asked in Top Tech Interviews
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