Node Matrix Validator 38 — 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 validator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Matrix Validator 38"
WHY DOES IT MATTER?
DP transforms exponential brute‑force into tractable linear time, essential for real‑time validation pipelines.
OPTIMIZATION CHALLENGE
The key is to identify a minimal constant‑size state that fully represents the prefix’s validity, cutting the combinatorial explosion.
REAL-WORLD CONNECTION
Think of streaming sensor data where each new reading must be validated against historical constraints without re‑scanning the whole log.
Start by writing the recurrence on paper, then collapse the recurrence to a rolling variable before coding.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
Dynamic programming (DP) solves problems that exhibit optimal substructure and overlapping subproblems by storing intermediate results, turning exponential recursions into polynomial time solutions. In the Node Matrix Validator, each prefix of the data sequence can be validated independently, and the result for a longer prefix depends only on a small, constant‑size state derived from the previous prefix, making DP the natural optimal paradigm.
A naive brute‑force scan that recomputes validation from scratch for every new element leads to O(N^2) or worse, because each element would re‑process the entire preceding subsequence. By defining a DP state that captures the necessary node‑matrix metrics (e.g., cumulative sums, parity, or max‑min constraints), we can update the state in O(1) per element, achieving linear overall complexity while guaranteeing correctness through memoization of the state transitions.
Interview Questions on This Problem
Q1Why is DP preferred over a simple iterative check for the Node Matrix Validator?
DP captures overlapping subproblems and avoids recomputing the same prefix validation repeatedly. It reduces the time from quadratic to linear while preserving correctness.
Q2What constitutes the DP state in this problem?
The state typically includes aggregated metrics such as cumulative sums, min/max values, or parity flags needed to validate the current node against the matrix constraints. It is a constant‑size tuple that can be updated in O(1).
Q3How can you further reduce the space complexity of the DP solution?
Since each transition depends only on the immediate previous state, you can overwrite the same variables instead of storing an entire array. This compresses space from O(N) to O(1).
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 25, 50]
Output
75
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 25, 50], we first sort the array in ascending order. Then, we iterate through the sorted array and add all numbers less than or equal to K to the sum. Since 50 is adjacent to numbers less than or equal to K, we also add 50 to the sum. The final sum is 75.
Input
[4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
Output
15
Explanation: Step-by-step: Given the input [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], we first sort the array in ascending order. Then, we iterate through the sorted array and add all numbers less than or equal to K to the sum. The final sum is 15.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a constant‑size DP state and update it iteratively, achieving O(N) time and O(1) space.
Brute Force Approach
Re‑evaluate the entire prefix for every new element, leading to O(N^2) time and O(1) extra space.
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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