Adaptive Cycle Metric — Problem Statement & Solution Guide
Problem Description
You are given an array or sequence of length $N$ representing numerical values or system metrics. Your task is to compute the adaptive cycle metric according to the target algorithm rules.
Formally, analyze the data sequence, process edge cases, and return the exact optimal result.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Adaptive Cycle Metric"
WHY DOES IT MATTER?
Rerooting DP transforms a seemingly quadratic problem into linear time, which is essential for any production system that processes massive hierarchical data (e.g., file‑system trees, organization charts, or network topologies). Mastery of this pattern unlocks a whole class of tree‑wide queries that would otherwise be intractable.
OPTIMIZATION CHALLENGE
The key insight is that the metric for a child can be expressed as the parent’s metric plus a correction term that depends only on the size (or weight) of the child’s sub‑tree. Recognizing this linear relationship enables O(1) per‑edge updates and eliminates redundant traversals.
REAL-WORLD CONNECTION
Consider a distributed monitoring system where each service node needs to know the total latency to all other services. Recomputing latencies from scratch for each node would be prohibitive; instead, the system propagates incremental updates as the logical root moves, mirroring the rerooting DP technique.
When coding the solution, first write a clean post‑order DFS that returns both subtree size and the metric for the chosen root. Then, in the pre‑order pass, pass the parent’s answer as a parameter; avoid global mutable state to keep the code bug‑free and easy to debug.
COMPLEXITY AT A GLANCE
O(N)O(N)Core Theory — Why This Approach?
The Adaptive Cycle Metric is a classic example of a tree‑wide DP problem where a value must be aggregated for every possible root of a tree. The naive way is to treat each node as the root, run a full DFS to compute the metric, and keep the best – this costs O(N^2) for N nodes because each DFS traverses O(N) edges. For large N (10^5‑10^6) such quadratic time is infeasible. The optimal paradigm uses a two‑pass rerooting DP: first compute the metric for an arbitrary root in a post‑order traversal, storing sub‑tree contributions; then, in a pre‑order pass, efficiently recompute the metric for each child by “moving” the root, adjusting the previously computed values with O(1) arithmetic per edge. This leverages the fact that the metric is a linear combination of sub‑tree aggregates, allowing constant‑time updates when the root shifts.
The second pass essentially propagates the answer from parent to child using the relation: answer[child] = answer[parent] + (globalAdjustment), where globalAdjustment depends on the size of the child’s sub‑tree and the total size of the tree. By maintaining subtree sizes and partial sums, we can compute the adaptive metric for every node in total O(N) time. The approach also extends to variants that require maximum/minimum values, weighted edges, or modular arithmetic, making rerooting DP a versatile tool for many hard tree problems.
Interview Questions on This Problem
Q1How would you compute the sum of distances from every node to all other nodes in a tree in O(N) time?
Perform a post‑order DFS to compute subtree sizes and the sum of distances for an arbitrary root. Then run a pre‑order DFS to reroot: for each child, sumDist[child] = sumDist[parent] + (N - 2*subtreeSize[child]). This updates the answer in O(1) per edge, yielding O(N) total.
Q2Explain why a naive O(N^2) solution fails for tree DP problems on N = 2·10^5 and how rerooting DP overcomes this.
A naive solution recomputes the whole metric for each possible root, leading to N traversals of O(N) each, which exceeds time limits (≈4·10^10 operations). Rerooting DP reuses previously computed sub‑tree information, updating the metric in constant time when moving the root across an edge, thus collapsing the total work to two linear traversals.
Q3In a weighted tree, how would you adapt the rerooting formula for the Adaptive Cycle Metric?
Store both the total weight of each sub‑tree and the weighted sum of distances. When moving the root from parent to child across edge weight w, adjust: answer[child] = answer[parent] + w * (N - 2*subtreeSize[child]) + (totalWeightParent - totalWeightChild) - totalWeightChild. The extra terms account for the edge weight and the shift in contribution of each side of the cut.
Examples
Input
[5, 2, 9, 6]
Output
22
Explanation: To compute the adaptive cycle metric, we need to analyze the data sequence. The given array is [5, 2, 9, 6]. The sum of the array is 5 + 2 + 9 + 6 = 22. Therefore, the adaptive cycle metric for this array is 22.
Input
[8, 6]
Output
14
Explanation: To compute the adaptive cycle metric, we need to analyze the data sequence. The given array is [8, 6]. The sum of the array is 8 + 6 = 14. Therefore, the adaptive cycle metric for this array is 14.
Constraints
- 1 <= N <= 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity expected: O(N) or O(N log N)
- Space Complexity expected: O(1) or O(N)
Optimal Approach & Strategy
Perform a post‑order DFS to compute subtree aggregates for an arbitrary root, then a pre‑order rerooting pass that updates the metric for each child in O(1) using the parent’s answer, achieving O(N) total time.
Brute Force Approach
For each node, run a full DFS/BFS to compute the metric, storing the best result; this repeats N times, leading to O(N^2) time.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int> nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
sum = 0
for num in nums:
sum += num
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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