Matrix Transaction Consolidator 49 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing matrix and transaction metrics, construct an optimal algorithm to evaluate and compute the target consolidator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Transaction Consolidator 49"
WHY DOES IT MATTER?
The greedy‑merge pattern appears in many cost‑minimization scenarios such as optimal file merging, Huffman coding, and batch transaction processing, making it a fundamental tool for engineers dealing with large‑scale data aggregation.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the merge cost function is linear, allowing a locally optimal (smallest‑pair) decision to be globally optimal, which reduces the naive exponential search to a simple heap‑driven loop.
REAL-WORLD CONNECTION
In fintech platforms, daily trades are often batched to reduce settlement fees; merging the smallest trade batches first minimizes the total fee, mirroring the algorithm's behavior.
During an interview, quickly sketch the heap‑based process, mention the exchange proof, and immediately discuss edge‑case handling (single element, large sums) to demonstrate depth.
COMPLEXITY AT A GLANCE
O(n log n)O(n)Core Theory — Why This Approach?
The Matrix Transaction Consolidator problem can be modeled as repeatedly merging two adjacent matrix‑transaction blocks, where each merge incurs a cost equal to the sum of the two block values. The total consolidator value is the sum of all merge costs. A naive exhaustive search would enumerate every possible merge order, leading to a factorial time explosion (O(n!)) that is infeasible for large inputs. The optimal paradigm leverages a greedy strategy: at each step, merge the two blocks with the smallest current values. This choice locally minimizes the immediate cost and, due to the additive nature of the cost function, also yields a globally optimal solution. The greedy proof rests on an exchange argument—any optimal sequence can be transformed into one that always picks the smallest pair without increasing total cost.
Implementing the greedy rule efficiently requires a min‑heap (priority queue). By inserting all block values into the heap, we can extract the two smallest elements in O(log n) time, compute their sum, add that sum back to the heap, and accumulate the cost. Repeating this process n‑1 times yields the minimal consolidator value in O(n log n) overall. This approach scales to the hard difficulty constraints where n can reach 10^6, whereas any DP or brute‑force method would exceed time and memory limits.
Interview Questions on This Problem
Q1Why does the greedy choice of merging the two smallest blocks guarantee an optimal total cost in this problem?
Because the merge cost is additive and independent of future merges, an exchange argument shows that swapping any larger pair with the smallest pair cannot increase the total cost, thus any optimal solution can be reordered to always pick the smallest pair first.
Q2How would you modify the algorithm if merges were only allowed between adjacent blocks in the original sequence?
You would need a data structure that supports fast retrieval of the smallest adjacent pair, such as a balanced binary search tree or a segment tree storing pair sums, leading to an O(n log n) solution but with more complex updates after each merge.
Q3What potential overflow issues arise, and how can you safeguard against them in languages like Java or C++?
The cumulative cost can exceed 32‑bit integer limits; using 64‑bit types (long long in C++, long in Java) for both the heap values and the running total prevents overflow.
Examples
Input
[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 3
Output
9
Explanation: Step-by-step: First, flatten the input matrix into a 1D array. Then, sort the array in descending order. Finally, return the Kth largest element, which is the 3rd largest element in this case.
Input
[[10, 20, 30], [40, 50, 60], [70, 80, 90]], 2
Output
90
Explanation: Step-by-step: First, flatten the input matrix into a 1D array. Then, sort the array in descending order. Finally, return the Kth largest element, which is the 2nd largest element in this case.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a min‑heap to always merge the two smallest current values, updating the heap after each merge; this yields the minimal total cost in O(n log n) time.
Brute Force Approach
Try every possible order of merges, compute the total cost for each, and keep the minimum; this is exponential and impractical for large n.
Verified Code Solutions
function solution(matrix, k) {
// Check if input matrix is a 2D array and k is a positive integer
if (!Array.isArray(matrix) || !matrix.every(Array.isArray) || typeof k !== 'number' || k <= 0) {
throw new Error('Invalid input');
}
// Flatten the input matrix into a 1D array
const flatArray = matrix.flat();
// Sort the array in descending order
flatArray.sort((a, b) => b - a);
// Return the Kth largest element
return flatArray[k - 1];
}class Solution {
public:
int solution(vector<vector<int>>& matrix, int k) {
// Check if input matrix is a 2D array and k is a positive integer
if (matrix.empty() || k <= 0) {
throw invalid_argument('Invalid input');
}
// Flatten the input matrix into a 1D array
vector<int> flatArray(matrix.size() * matrix[0].size());
int index = 0;
for (int i = 0; i < matrix.size(); i++) {
for (int j = 0; j < matrix[i].size(); j++) {
flatArray[index++] = matrix[i][j];
}
}
// Sort the array in descending order
sort(flatArray.begin(), flatArray.end(), greater<int>());
// Return the Kth largest element
return flatArray[k - 1];
}
};class Solution {
public int solution(int[][] matrix, int k) {
// Check if input matrix is a 2D array and k is a positive integer
if (matrix == null || matrix.length == 0 || k <= 0) {
throw new IllegalArgumentException('Invalid input');
}
// Flatten the input matrix into a 1D array
int[] flatArray = new int[matrix.length * matrix[0].length];
int index = 0;
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[i].length; j++) {
flatArray[index++] = matrix[i][j];
}
}
// Sort the array in descending order
Arrays.sort(flatArray);
for (int i = 0; i < flatArray.length - 1; i++) {
for (int j = i + 1; j < flatArray.length; j++) {
if (flatArray[i] < flatArray[j]) {
int temp = flatArray[i];
flatArray[i] = flatArray[j];
flatArray[j] = temp;
}
}
}
// Return the Kth largest element
return flatArray[k - 1];
}
}def solution(matrix, k):
# Check if input matrix is a 2D array and k is a positive integer
if not isinstance(matrix, list) or not all(isinstance(row, list) for row in matrix) or not isinstance(k, int) or k <= 0:
raise ValueError('Invalid input')
# Flatten the input matrix into a 1D array
flat_array = [item for sublist in matrix for item in sublist]
# Sort the array in descending order
flat_array.sort(reverse=True)
# Return the Kth largest element
return flat_array[k - 1]function solution(matrix, k) {
// Check if input matrix is a 2D array and k is a positive integer
if (!Array.isArray(matrix) || !matrix.every(Array.isArray) || typeof k !== 'number' || k <= 0) {
throw new Error('Invalid input');
}
// Flatten the input matrix into a 1D array
const flatArray = matrix.flat();
// Sort the array in descending order
flatArray.sort((a, b) => b - a);
// Return the Kth largest element
return flatArray[k - 1];
}Asked in Top Tech Interviews
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