Node Matrix Partition 6 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing node and matrix metrics, construct an optimal algorithm to evaluate and compute the target partition value under given operational constraints, where K is the threshold value and the goal is to select the maximum sum of K components greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Matrix Partition 6"
WHY DOES IT MATTER?
Sliding‑window patterns turn quadratic‑time brute‑force scans into linear passes, which is essential for real‑time analytics, streaming data, and any scenario where input size can reach millions or billions.
OPTIMIZATION CHALLENGE
The key insight is recognizing overlapping sub‑problems: each window shares K‑1 elements with its neighbor, allowing O(1) incremental updates instead of recomputing the whole sum.
REAL-WORLD CONNECTION
Think of a network router that monitors the total traffic over the last K seconds to trigger alerts; the router updates the traffic count by subtracting the oldest second and adding the newest, exactly mirroring the sliding‑window update.
During an interview, compute the first window sum explicitly, then write a single loop that slides the window while updating the sum and tracking the maximum; avoid extra arrays or nested loops.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem is a classic sliding‑window scenario where we must examine all contiguous blocks of exactly K elements in a linear sequence and compute the block with the highest aggregate metric. A naïve solution would recompute the sum for each window from scratch, leading to O(N·K) time, which quickly becomes prohibitive when N and K are large (e.g., N up to 10^6). The optimal paradigm leverages the fact that consecutive windows overlap by K‑1 elements: when the window slides one position to the right, we subtract the element that exits the window and add the new entrant. This constant‑time update yields an O(N) overall runtime while using O(1) auxiliary space. The sliding‑window technique is a deterministic, greedy approach that guarantees the global optimum for fixed‑size windows because the sum of each window is independent of future choices; we simply need to keep track of the maximum observed sum that also satisfies the constraint sum > K.
Interview Questions on This Problem
Q1How would you modify the sliding‑window solution if the window size K is not fixed but you need the maximum sum of any subarray whose length is at most K?
Maintain a deque that stores prefix sums; for each index i compute prefix[i] and query the smallest prefix within the last K positions to maximize prefix[i] - minPrefix. This yields O(N) time and O(K) space.
Q2Explain why a binary‑search on the answer (maximum sum) combined with a check function can also solve the problem, and compare its complexity to the sliding‑window approach.
We can binary‑search the possible sum range and, for each candidate S, verify if any window of size K has sum ≥ S using a sliding window in O(N). The overall complexity becomes O(N·log M) where M is the sum range, which is slower than the direct O(N) method but useful when the window size is variable or additional constraints exist.
Q3In a distributed system where the data stream is sharded across multiple nodes, how would you compute the global maximum K‑window sum efficiently?
Each shard computes its local maximum K‑window sum and also retains the prefix sum of its first K‑1 elements and suffix sum of its last K‑1 elements. A coordinator then merges these border contributions to evaluate windows that span shards, achieving O(totalElements) work with minimal cross‑node communication.
Examples
Input
[60, 70, 80, 90, 50, 35, 30, 25, 20, 15]
Output
150
Explanation: Step-by-step: Given the input array, first sort it in descending order. Then, iterate through the sorted array and add the K largest components greater than K to the sum. For this example, the sum of the 5 largest components greater than K is 150 (60 + 70 + 80 + 90 + 50).
Input
[35, 30, 25, 20, 15, 10, 5, 1]
Output
60
Explanation: Step-by-step: Given the input array, first sort it in descending order. Then, iterate through the sorted array and add the K largest components greater than K to the sum. For this example, the sum of the 3 largest components greater than K is 60 (35 + 30 + 25).
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a sliding window: compute the first K‑sum, then update it in O(1) as the window moves, achieving O(N) time and O(1) extra space.
Brute Force Approach
Re‑calculate the sum of every possible K‑length subarray from scratch, leading to O(N·K) time.
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > k && k > 0) {
sum += nums[i];
k--;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.begin(), nums.end(), greater<int>());
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > k && k > 0) {
sum += nums[i];
k--;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > k && k > 0) {
sum += nums[i];
k--;
}
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
sum = 0
for num in nums:
if num > k and k > 0:
sum += num
k -= 1
return sumfunction solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > k && k > 0) {
sum += nums[i];
k--;
}
}
return sum;
}Asked in Top Tech Interviews
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