Payload Token Evaluator 15 — 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 evaluator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Token Evaluator 15"
WHY DOES IT MATTER?
Post‑order traversal guarantees that child values are available before computing a parent’s value, which is essential for dynamic programming on trees. Without this order, you would need to recompute subtrees or store intermediate results in a separate structure, increasing complexity.
OPTIMIZATION CHALLENGE
The critical insight is to compute each node’s evaluator value once and propagate it upward, reducing the problem from exponential to linear time. This eliminates redundant recomputation of identical subtrees.
REAL-WORLD CONNECTION
In distributed build systems, a file’s build artifact depends on its dependencies. Evaluating the build order is analogous to a post‑order traversal where each file’s artifact is computed after all its dependencies are built.
When implementing the DFS, use an iterative stack or tail‑recursion to avoid stack overflow on deep trees, and consider using a HashMap to memoize results if the tree contains shared substructures.
COMPLEXITY AT A GLANCE
O(n)O(h)Core Theory — Why This Approach?
The Payload Token Evaluator problem reduces to computing a value that depends on the payload and token metrics of each node in a binary tree. A naive approach would recursively evaluate each node by recomputing the values of its subtrees for every query, leading to exponential time complexity because the same subtrees are processed repeatedly. In contrast, the optimal paradigm performs a single depth‑first traversal (post‑order) where each node’s value is computed once from its children’s already‑computed results. This transforms the problem into a linear‑time dynamic programming on trees, ensuring that each node is visited only once and that intermediate results are reused efficiently.
The key insight is that the evaluator value at a node is a deterministic function of its own payload, token, and the evaluator values of its left and right children. By propagating these values upward in a bottom‑up fashion, we avoid redundant work and achieve optimal time and space usage. This pattern is a classic example of tree DP, where local computations are combined to form a global solution.
Moreover, the problem’s constraints often involve large trees (up to 10^5 nodes) and strict time limits, making any approach that touches each node more than a constant number of times infeasible. Using iterative stack or tail‑recursive techniques can further reduce stack overhead, making the solution robust for deep trees that would otherwise cause stack overflow in a naive recursive implementation.
Interview Questions on This Problem
Q1How would you compute the maximum payload sum of any root‑to‑leaf path in a binary tree?
I would perform a depth‑first search, passing along the cumulative sum from the root to the current node. At each leaf, I compare the cumulative sum to a global maximum and update it if larger. This yields an O(n) time solution with O(h) space for recursion depth.
Q2Explain how to evaluate an arithmetic expression represented as a binary tree.
I would use a post‑order traversal: evaluate the left subtree, then the right subtree, and finally apply the operator at the current node to the two results. This ensures that operands are computed before the operator, giving an O(n) time and O(h) space solution.
Q3Describe how to find the lowest common ancestor of two nodes in a binary tree without parent pointers.
I would recursively search for the two target nodes. If both left and right subtrees return non‑null, the current node is the LCA. If only one side returns non‑null, propagate that node upward. This runs in O(n) time and O(h) space.
Examples
Input
[10, 20, 30, 40, 50], 40
Output
90
Explanation: Step-by-step: Given the input [10, 20, 30, 40, 50] and K = 40, we first filter out numbers greater than K, which are 50. Then, we sum up the remaining numbers, 40 and 30, and 20 and 10. The sum is 90.
Input
[10, 20, 30, 40, 50], 60
Output
0
Explanation: Step-by-step: Given the input [10, 20, 30, 40, 50] and K = 60, we first filter out numbers greater than K, which are all numbers in the array. Since there are no numbers left, the sum is 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
The optimal solution performs a single post‑order DFS, computing each node’s evaluator value once from its children’s results, achieving linear time and linear space for recursion depth.
Brute Force Approach
A naive solution would recursively recompute the evaluator value for each node’s subtrees for every query, leading to exponential time as the same subtrees are processed repeatedly.
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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