Node Matrix Evaluator 25 — 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 evaluator value under given operational constraints. The input array nums and the integer K are provided.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Matrix Evaluator 25"
WHY DOES IT MATTER?
Sliding windows turn quadratic scans into linear passes, crucial for high‑throughput data streams.
OPTIMIZATION CHALLENGE
The key is to update the aggregate in O(1) while the window slides one element.
REAL-WORLD CONNECTION
Network routers compute moving averages of packet rates using the same principle.
Initialize the first window fully, then reuse the same variables to avoid extra allocations.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The sliding window technique transforms a naïve O(N·K) enumeration of all length‑K subarrays into a linear scan by maintaining a running sum of the current window. As the window slides one step, we subtract the element exiting the window and add the new entrant, preserving O(1) update cost.
On large inputs, the naïve double loop incurs excessive time and cache misses, while the optimal paradigm leverages the overlapping nature of consecutive subarrays, guaranteeing O(N) time and O(1) auxiliary space, which is essential for real‑time or memory‑constrained environments.
Interview Questions on This Problem
Q1How does the sliding window reduce the time complexity compared to the brute‑force method?
It reuses the sum of the previous window, updating it in constant time instead of recomputing from scratch. This changes the overall complexity from O(N·K) to O(N).
Q2What edge case must you handle when K equals the length of the array?
The window never slides, so the answer is simply the sum of the entire array. Ensure you don’t attempt to access out‑of‑bounds indices.
Q3Can the sliding window be applied to non‑contiguous subarray problems?
Only when subproblems exhibit a fixed-size, overlapping structure. For arbitrary subsets, other techniques like prefix sums or DP are needed.
Examples
Input
[1, 2, 3, 4, 5], 5
Output
15
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5] and K = 5, we first sort the array in ascending order. Then, we iterate through the sorted array and sum up all the numbers less than or equal to K, which are 1, 2, 3, 4, and 5. Therefore, the output is 15.
Input
[10, 20, 30, 40, 50], 25
Output
25
Explanation: Step-by-step: Given the input [10, 20, 30, 40, 50] and K = 25, we first sort the array in ascending order. Then, we iterate through the sorted array and sum up all the numbers less than or equal to K, which is 25. Therefore, the output is 25.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Compute the sum of the first K elements, then slide the window, updating the sum in O(1) per step while tracking the max.
Brute Force Approach
Iterate over every possible start index, compute the sum of the next K elements each time, and keep the maximum.
Verified Code Solutions
function solution(nums, K) {
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] <= K) {
sum += nums[i];
} else {
break;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
sort(nums.begin(), nums.end());
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] <= K) {
sum += nums[i];
} else {
break;
}
}
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) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
}def solution(nums, K):
nums.sort()
sum = 0
for num in nums:
if num <= K:
sum += num
else:
break
return sumfunction solution(nums, K) {
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] <= K) {
sum += nums[i];
} else {
break;
}
}
return sum;
}Asked in Top Tech Interviews
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