Payload Token Detector 48 — Problem Statement & Solution Guide
Problem Description
You are tasked with implementing a deep inspection routine for a hierarchical data structure representing nested payload segments. The structure is defined as a tree where each node contains an integer weight and a list of child nodes. The goal is to compute the 'Detector Value' for the entire structure. The Detector Value is defined as the sum of the weights of all nodes that are reachable via a Depth-First Search (DFS) traversal, subject to a specific pruning constraint: any subtree rooted at a node with a weight of 0 is considered 'corrupted' and must be entirely excluded from the calculation, including the root node itself and all its descendants. If the root node has a weight of 0, the result is 0. Otherwise, traverse the tree using DFS, accumulating the weights of valid nodes while skipping any branch that encounters a zero-weight node.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Token Detector 48"
WHY DOES IT MATTER?
Iterative DFS with a stack turns a potentially exponential recursion into a linear, safe traversal.
OPTIMIZATION CHALLENGE
The key is to compute depth on the fly to avoid extra passes or recomputation.
REAL-WORLD CONNECTION
Similar patterns appear in parsing nested JSON or XML payloads where depth matters for weighting.
Push children in reverse order if order matters, and reuse a single struct to keep memory footprints low.
COMPLEXITY AT A GLANCE
O(n)O(h)Core Theory — Why This Approach?
The Detector Value is the weighted depth sum of a rooted tree: for each node, multiply its weight by its depth (root depth = 1) and accumulate. An iterative depth‑first traversal using an explicit stack preserves O(h) auxiliary space, where h is the tree height, and avoids the call‑stack overflow risk of recursive DFS on deep or skewed trees. Naïve recursion recomputes depth on each return or uses global counters, leading to O(n h) time in pathological cases and stack overflow for n > 10⁵. The optimal paradigm is a single-pass stack‑driven DFS that pushes child nodes together with their computed depth, guaranteeing linear O(n) time and O(h) space regardless of tree shape.
Interview Questions on This Problem
Q1Why is an explicit stack preferred over recursion for deep trees?
Recursion relies on the language call stack, which may overflow for depth > typical limits. An explicit stack gives controlled memory usage and works uniformly across languages.
Q2How does storing depth alongside each node in the stack simplify the algorithm?
It eliminates the need for back‑tracking or separate depth counters, allowing each node's contribution to be computed immediately. This yields a clean O(1) per‑node operation.
Q3What is the overall time and space complexity of the stack‑based solution?
The algorithm visits each node once, so time is O(n). The stack holds at most the height of the tree, giving O(h) auxiliary space.
Examples
Input
root = Node(5, [Node(3, []), Node(0, [Node(10, [])])])
Output
8
Explanation: Start at root (5). Weight is non-zero, add 5 to sum. Sum = 5. Visit first child (3). Weight is non-zero, add 3 to sum. Sum = 8. No children. Visit second child (0). Weight is zero. Prune this entire subtree. Do not add 0 or its child (10). Final sum is 8.
Input
root = Node(0, [Node(1, []), Node(2, [])])
Output
0
Explanation: Start at root (0). Weight is zero. The entire tree is corrupted. Prune immediately. No nodes are added to the sum. Final sum is 0.
Input
root = Node(10, [Node(2, [Node(5, [])]), Node(4, [])])
Output
21
Explanation: Start at root (10). Add 10. Sum = 10. Visit first child (2). Add 2. Sum = 12. Visit its child (5). Add 5. Sum = 17. Visit second child of root (4). Add 4. Sum = 21. All nodes are valid. Final sum is 21.
Input
root = Node(1, [Node(2, [Node(0, [Node(100, [])])]), Node(3, [])])
Output
6
Explanation: Start at root (1). Add 1. Sum = 1. Visit first child (2). Add 2. Sum = 3. Visit its child (0). Weight is zero. Prune subtree. Do not add 0 or 100. Visit second child of root (3). Add 3. Sum = 6. Final sum is 6.
Constraints
- 1 <= number of nodes <= 10^5
- 0 <= node.weight <= 10^9
- The tree is guaranteed to be acyclic (a valid tree structure).
- The depth of the tree may be up to 10^5, requiring an iterative DFS or increased recursion limit to avoid stack overflow.
Optimal Approach & Strategy
Iterative DFS with a stack that stores (node, depth) pairs, updating the sum in O(1) per node.
Brute Force Approach
Recursively compute depth for each node on every visit, leading to repeated traversals and possible stack overflow.
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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