Network Node Consolidator 43 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing network and node metrics, construct an optimal algorithm to evaluate and compute the target consolidator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Node Consolidator 43"
WHY DOES IT MATTER?
The "maximum sum of non‑adjacent elements" pattern appears whenever a decision at position i influences the feasibility of decisions at i+1, a common scenario in resource allocation, scheduling, and financial portfolio selection.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that only the two previous optimal values are needed to decide the current state, collapsing an O(n) DP table into two scalar variables and thus achieving O(1) auxiliary space.
REAL-WORLD CONNECTION
Think of a data‑center where activating a node consumes power and creates heat; you cannot turn on two neighboring racks simultaneously. The DP computes the maximal throughput while respecting the cooling constraint.
During an interview, write the recurrence first, then immediately translate it into the rolling‑variable implementation – this shows you understand both the theory and the practical space optimization.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem belongs to the family of linear‑state dynamic programming where the optimal solution for a prefix of the input can be expressed in terms of optimal solutions of smaller prefixes. By defining a state that captures the best consolidator value up to index i, we can decide whether to include the i‑th element (and thus add its metric to the best value at i‑2) or skip it (and inherit the best value at i‑1). Naïve recursion explores every subset of indices, leading to an exponential 2^n search space because each element has a binary choice – include or exclude – and overlapping sub‑problems are recomputed many times. The DP paradigm eliminates this redundancy by storing intermediate results, turning the exponential blow‑up into a linear scan. The optimal recurrence is: dp[i] = max(dp[i‑1], dp[i‑2] + metric[i]), with dp[0] = metric[0] and dp[1] = max(metric[0], metric[1]). This yields an O(n) time algorithm with O(1) extra space when we keep only the last two values.
Interview Questions on This Problem
Q1How would you modify the DP solution if the constraint changes from "no two adjacent nodes" to "no three consecutive nodes can be selected"?
Introduce a three‑state DP: dp[i][0] – best value ending at i without picking i, dp[i][1] – best value picking i but not i‑1, dp[i][2] – best value picking i and i‑1. The recurrence updates each state from the previous three states while ensuring the three‑consecutive rule, still O(n) time and O(1) space.
Q2Explain why a greedy approach that always picks the largest remaining metric fails for this problem.
Greedy selection ignores the adjacency constraint; picking the global maximum may block two high‑value neighbors that together yield a larger total. Counter‑example: metrics = [4, 5, 4]; greedy picks 5, total = 5, while optimal picks 4 + 4 = 8.
Q3In a distributed system, how could you compute the consolidator value in parallel across shards of the sequence?
Each shard computes three local aggregates: best value when the first element is taken, when it is skipped, and the maximum overall. A reduction step merges adjacent shard results using the same three‑state transition, preserving correctness while achieving O(log k) merge steps for k shards.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 5
Output
15
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and the threshold K = 5, we iterate through the array and sum up all numbers that are less than or equal to K. In this case, the numbers 1, 2, 3, 4, and 5 are less than or equal to K, so the sum is 1 + 2 + 3 + 4 + 5 = 15.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 10
Output
55
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and the threshold K = 10, we iterate through the array and sum up all numbers that are less than or equal to K. In this case, all numbers in the array are less than or equal to K, so the sum is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Iterate once, maintaining two variables that store the best value up to the previous and the one before that, updating them with the DP recurrence.
Brute Force Approach
Recursively explore every subset of indices, deciding for each element to include it (if the previous wasn't taken) or skip it, which leads to exponential time.
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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