Matrix Vessel Aligner 12 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing matrix and vessel metrics, construct an optimal algorithm to evaluate and compute the target aligner value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Vessel Aligner 12"
WHY DOES IT MATTER?
DP turns an intractable combinatorial explosion into a tractable polynomial solution.
OPTIMIZATION CHALLENGE
The key is collapsing overlapping sub‑problems into a compact state space to cut time from exponential to O(n·k).
REAL-WORLD CONNECTION
It mirrors resource allocation in data‑center load balancers where tasks (matrix cells) must be assigned to limited servers (vessels) efficiently.
Always pre‑compute prefix aggregates (sums, mins) to make each transition O(1) and keep the inner loop tight.
COMPLEXITY AT A GLANCE
O(n·k)O(n·k)Core Theory — Why This Approach?
Dynamic programming solves the Matrix Vessel Aligner problem by exploiting overlapping sub‑problems: each prefix of the input sequence can be aligned optimally based on previously computed states, turning an exponential search into a polynomial one. The naive recursive enumeration tries every possible partition of the sequence, leading to O(2^n) or worse time, which quickly explodes for n > 30.
The optimal paradigm defines a DP state dp[i][j] (or a 1‑D reduction) representing the best aligner value for the first i elements with j vessels used, and transitions by adding the i‑th element to an existing vessel or starting a new one while respecting constraints. By memoizing these transitions and pruning infeasible states, the algorithm runs in O(n·k) time (k = number of vessels) and O(n·k) space, achieving the required scalability for large inputs.
Interview Questions on This Problem
Q1How does DP reduce the exponential blow‑up in the Matrix Vessel Aligner problem?
DP stores the optimal result for each prefix and vessel count, avoiding recomputation of identical sub‑problems. This transforms the search space from exponential to polynomial.
Q2What is the typical DP state definition for this problem and why?
dp[i][j] = best aligner value using the first i elements with exactly j vessels, because the decision for element i only depends on the previous prefix and vessel count. It captures all necessary information for optimal substructure.
Q3When can the DP be reduced from 2‑D to 1‑D, and what trade‑off does it introduce?
If transitions only depend on the previous row (i‑1), we can roll the array into a 1‑D vector, saving space. The trade‑off is careful reverse iteration to prevent overwriting needed values.
Examples
Input
[100, 90, 80, 70, 60, 50, 40, 30, 20], 50
Output
210
Explanation: Step-by-step: Given the input array [100, 90, 80, 70, 60, 50, 40, 30, 20] and K = 50, we first sort the array in descending order. Then, we initialize a variable sum to 0 and iterate over the sorted array. We add each element to the sum if it is strictly greater than K. The first 3 elements greater than K are 60, 70, and 80, so the correct output is 60 + 70 + 80 = 210.
Input
[45, 35, 25, 15, 5], 35
Output
45
Explanation: Step-by-step: Given the input array [45, 35, 25, 15, 5] and K = 35, we first sort the array in descending order. Then, we initialize a variable sum to 0 and iterate over the sorted array. We add each element to the sum if it is strictly greater than K. The first element greater than K is 45, so the correct output is 45.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use DP with state dp[i][j] and O(1) transition using pre‑computed prefix metrics, optionally rolling to 1‑D for space efficiency.
Brute Force Approach
Enumerate every possible way to split the sequence into up to k vessels and compute the score for each partition.
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) {
sum += nums[i];
if (i === 2) break;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
sort(nums.rbegin(), nums.rend());
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > K) {
sum += nums[i];
if (i == 2) 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];
if (i == 2) break;
}
}
return sum;
}
}def solution(nums, K):
nums.sort(reverse=True)
sum = 0
for i in range(len(nums)):
if nums[i] > K:
sum += nums[i]
if i == 2: break
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) {
sum += nums[i];
if (i === 2) break;
}
}
return sum;
}Asked in Top Tech Interviews
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