Dynamic Node Cluster — Problem Statement & Solution Guide
Problem Description
Dynamic Node Cluster
You are given a list of integer values that represent the capacities of nodes in a distributed system. Your task is to determine the smallest even capacity that is at least as large as a specified target value. If no such even capacity exists in the list, report that the requirement cannot be met.
Input format:
- The first line contains an integer N (1 ≤ N ≤ 10^5), the number of node capacities.
- The second line contains N space‑separated integers, each representing a node capacity.
- The third line contains a single integer T, the target capacity.
Output format:
- Output a single integer: the minimal even capacity that is greater than or equal to T, or -1 if no such capacity exists.
The solution should be efficient, using binary search on a sorted list of the even capacities to achieve O(N log N) time complexity.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Dynamic Node Cluster"
WHY DOES IT MATTER?
Lower‑bound binary search on a sorted array is a fundamental pattern for "first element satisfying a predicate" problems. Mastering it lets you turn many linear‑time scans into logarithmic queries after a one‑time sort.
OPTIMIZATION CHALLENGE
The key insight is to reduce the search space before applying binary search: filter out non‑even values so the predicate "value is even" is already satisfied, allowing the binary search to focus solely on the magnitude condition.
REAL-WORLD CONNECTION
In distributed systems, node capacities are often indexed in a monitoring database. When allocating a new workload, the scheduler needs the smallest node that can accommodate the load while meeting parity constraints (e.g., even‑sized memory blocks). Sorting the capacity table and binary searching mirrors how such schedulers quickly locate a suitable machine.
During an interview, sort the evens first, then write a clean lower‑bound routine (or use language‑provided functions). Keep edge‑case handling (no answer) separate from the search logic to avoid off‑by‑one bugs.
COMPLEXITY AT A GLANCE
O(N log N) preprocessing + O(log N) per queryO(N) for the filtered arrayCore Theory — Why This Approach?
The problem reduces to a classic "search for the lower bound" scenario. By extracting only the even capacities and sorting them, we obtain a monotonic sequence where each element is greater than or equal to the previous one. In such a sorted array, binary search can locate the first element that is not smaller than the target in O(log N) time, guaranteeing the smallest qualifying even capacity. A naïve linear scan would examine every element, which is acceptable for a single query but becomes prohibitive when the same list must answer many target queries or when N approaches 10^5, because O(N) per query can lead to time‑outs. The optimal paradigm therefore combines preprocessing (filter + sort) with a logarithmic lower‑bound search, leveraging the binary search invariant that the search interval always contains the answer if it exists.
Interview Questions on This Problem
Q1How would you modify the solution if the list must support multiple target queries efficiently?
Preprocess once by filtering evens and sorting them. For each query, perform a binary search (lower_bound) to find the smallest even ≥ target. This yields O(N log N) preprocessing and O(log N) per query.
Q2What changes are needed if the requirement is to find the smallest odd capacity ≥ target?
The algorithm stays the same; only the filtering step changes to keep odd numbers instead of evens before sorting and binary searching.
Q3Can you achieve O(N) total time without sorting? Explain the trade‑offs.
Yes, by scanning once while tracking the minimum even ≥ target. This works for a single query but cannot answer subsequent queries without re‑scanning, so it sacrifices reusability for constant‑time per query after the first scan.
Examples
Input
6 3 8 5 12 7 10 9
Output
10
Explanation: The even capacities are 8, 12, and 10. The smallest even number that is ≥9 is 10.
Input
5 1 3 5 7 9 4
Output
-1
Explanation: There are no even capacities in the list, so the requirement cannot be satisfied.
Input
7 2 4 6 8 10 12 14 11
Output
12
Explanation: All capacities are even. The smallest even number ≥11 is 12.
Input
4 -2 -4 0 2 -3
Output
-2
Explanation: Even capacities are -4, -2, 0, 2. The smallest even number ≥-3 is -2.
Input
3 1000000000 999999998 999999996 999999997
Output
1000000000
Explanation: Even capacities are 1000000000, 999999998, 999999996. The smallest even number ≥999999997 is 1000000000.
Constraints
- 1 <= N <= 100000
- -1000000000 <= nums[i] <= 1000000000
- -1000000000 <= T <= 1000000000
Optimal Approach & Strategy
Filter evens, sort them, then perform a binary‑search lower‑bound to locate the first element ≥ target.
Brute Force Approach
Scan the entire list, keep track of the minimum even value that is ≥ target, and return it after the loop.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
let sum = 0;
for (let num of nums) {
if (num % 2 === 0) sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int> nums) {
if (nums.size() == 0) return 0;
int sum = 0;
for (int num : nums) {
if (num % 2 == 0) sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return 0;
int sum = 0;
for (int num : nums) {
if (num % 2 == 0) sum += num;
}
return sum;
}
}def solution(nums):
if not nums:
return 0
return sum(num for num in nums if num % 2 == 0)function solution(nums) {
if (nums.length === 0) return 0;
let sum = 0;
for (let num of nums) {
if (num % 2 === 0) sum += num;
}
return sum;
}Asked in Top Tech Interviews
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