Pipeline Vector Optimizer 33 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and vector metrics, construct an optimal algorithm to evaluate and compute the target optimizer value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Vector Optimizer 33"
WHY DOES IT MATTER?
The stateful DP pattern is essential because it transforms an exponential search over skip positions into a linear traversal. By encoding the skip state in the DP, we avoid redundant recomputation and capture all possibilities in a single pass.
OPTIMIZATION CHALLENGE
The core insight is that the skip can be applied at any node, so we must maintain two parallel best sums per node. This reduces the problem from exploring all skip positions to a constant‑time merge of child states.
REAL-WORLD CONNECTION
Consider a data processing pipeline where one stage can be optionally skipped to save resources. Deciding whether to skip that stage at each point in the pipeline is analogous to the DP state: "skip used" vs. "skip unused". The algorithm mirrors the decision logic in such pipelines.
When implementing, remember to return a pair (withoutSkip, withSkip) from each recursive call. Use a helper struct or tuple to keep the code clean, and always handle null children by returning (0, -∞) to avoid accidental skips.
COMPLEXITY AT A GLANCE
O(n)O(h)Core Theory — Why This Approach?
The problem reduces to a classic tree dynamic programming (DP) scenario where each node’s optimal value depends on its children’s optimal values. A naive approach would enumerate all root‑to‑leaf paths and, for each path, try every possible single‑node skip, leading to an exponential blow‑up (O(n·2^h) in the worst case). Instead, we use a bottom‑up DP that, for every node, keeps two pieces of information: the best sum achievable on a path that *has not yet used* the skip, and the best sum on a path that *has already used* the skip. By combining these two values from the left and right subtrees in constant time, we propagate optimal solutions up to the root in linear time. This pattern is essentially a “stateful DP on trees” where the state encodes whether a special operation (the skip) has been consumed.
The key insight is that the skip can be applied at any node along the path, so the DP must consider both possibilities at each step. By storing the best sums for both states, we avoid recomputing subproblems and eliminate the combinatorial explosion. The resulting algorithm runs in O(n) time and O(h) space (due to recursion stack), which is optimal for a tree of n nodes.
Naive recursion that recomputes the same sub‑paths for each skip choice would quickly become infeasible for large trees (e.g., 10^5 nodes). The DP approach guarantees that each node is processed once, and the state transition is constant time, making it scalable and suitable for interview scenarios that test both algorithmic thinking and efficient coding.
Interview Questions on This Problem
Q1How would you modify the classic maximum root‑to‑leaf path sum algorithm to allow skipping exactly one node on the path?
I would use a DP that tracks two values per node: the best sum without using the skip and the best sum after the skip has been used. For each child, I compute both values and then combine them: the best without skip is node.val + max(childWithoutSkip, childWithSkip), and the best with skip is max(childWithoutSkip, childWithSkip) (if we skip the current node) or node.val + max(childWithSkip). This yields an O(n) solution.
Q2What is the time and space complexity of your solution, and why is it optimal?
The time complexity is O(n) because each node is visited once and constant work is done per node. The space complexity is O(h) due to the recursion stack, where h is the tree height; in the worst case h = n, but for a balanced tree h = log n. This is optimal because any algorithm must at least read all nodes, which takes O(n) time.
Q3Can you explain a real‑world scenario where this “skip one node” optimization would be useful?
In distributed systems, a pipeline might have a single optional cache or a redundant node that can be bypassed to reduce latency. The algorithm models the decision to skip that node to maximize throughput, analogous to choosing whether to route traffic through a slower but more reliable node or to bypass it for speed.
Examples
Input
[30, 40, 25, 50, 60, 20, 10, 5]
Output
2
Explanation: Step-by-step: Given the input [30, 40, 25, 50, 60, 20, 10, 5], we first initialize a counter variable to 0. Then, we iterate through each element in the array. If the current element is greater than K (25 in this case), we increment the counter by 1. After iterating through all elements, the counter will hold the count of elements greater than K, which is 2 in this case.
Input
[10, 20, 30, 40, 50, 60]
Output
0
Explanation: Step-by-step: Given the input [10, 20, 30, 40, 50, 60], we first initialize a counter variable to 0. Then, we iterate through each element in the array. Since all elements are less than or equal to K (60 in this case), the counter remains 0 after iterating through all elements.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a post‑order traversal that returns two values per node: best sum without skip and best sum with skip already used. Combine child values in constant time to propagate optimal sums up to the root, achieving O(n) time.
Brute Force Approach
Enumerate every root‑to‑leaf path, then for each path try skipping each node one by one and compute the sum. This leads to exponential time complexity, O(n·2^h).
Verified Code Solutions
function solution(nums, K) {
let count = 0;
for (let num of nums) {
if (num > K) {
count++;
}
}
return count;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int count = 0;
for (int num : nums) {
if (num > K) {
count++;
}
}
return count;
}
};class Solution {
public int solution(int[] nums, int K) {
int count = 0;
for (int num : nums) {
if (num > K) {
count++;
}
}
return count;
}
}def solution(nums, K):
count = 0
for num in nums:
if num > K:
count += 1
return countfunction solution(nums, K) {
let count = 0;
for (let num of nums) {
if (num > K) {
count++;
}
}
return count;
}Asked in Top Tech Interviews
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