Matrix Transaction Consolidator 14 — 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 14"
WHY DOES IT MATTER?
Stacks turn quadratic range problems into linear scans.
OPTIMIZATION CHALLENGE
Identifying and maintaining the monotonic invariant reduces each element's work to constant time.
REAL-WORLD CONNECTION
Similar to order‑book matching engines that need nearest price levels instantly.
Push only when the invariant holds; pop aggressively to expose the next valid boundary.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The naive solution treats each matrix element and transaction metric independently, scanning all possible sub‑matrices or transaction windows to compute the consolidator value. This results in O(n^2) or worse time complexity because every element may be combined with every other, which quickly exceeds limits for large inputs.\nThe optimal paradigm leverages a monotonic stack to maintain a decreasing (or increasing) sequence of relevant metrics while iterating through the flattened matrix representation. By using the stack to instantly locate the nearest element that violates the transaction constraint, we can compute contributions of each element in O(1) amortized time, collapsing the overall algorithm to linear O(n) time and O(n) auxiliary space.
Interview Questions on This Problem
Q1How does a monotonic stack help reduce the time complexity for range‑based consolidator calculations?
It keeps elements in sorted order so the next greater or smaller boundary is found in O(1) amortized time. This eliminates the need for nested loops that examine every possible range.
Q2What is the key invariant maintained by the stack during the matrix traversal?
The stack stores indices of elements in strictly decreasing transaction metric order. Any element popped violates the invariant, indicating a boundary for the current element's contribution.
Q3Why must we process the matrix in a linearized (row‑major) order for this stack technique?
Linearization preserves the relative ordering required for correct nearest‑boundary detection. It also allows the stack to operate on a one‑dimensional sequence, simplifying implementation.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250, 260]
Output
2210
Explanation: Step-by-step: Given the input array [10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250, 260], we need to find the target consolidator value. We start by iterating through the array and summing up all elements greater than 150. In the first iteration, we sum 160 + 170 + 180 + 190 + 200 + 210 + 220 + 230 = 1610. In the second iteration, we sum 240 + 250 + 260 = 750. Adding these two sums together, we get 1610 + 750 = 2360. However, this is incorrect because we need to include the elements 110 and 120 in the sum. Therefore, the correct output is 2210.
Input
[150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250, 260]
Output
260
Explanation: Step-by-step: Given the input array [150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250, 260], we need to find the target consolidator value. We start by iterating through the array and summing up all elements greater than 150. In the first iteration, we sum 160 + 170 + 180 + 190 + 200 + 210 + 220 + 230 + 240 + 250 + 260 = 2600. However, this is incorrect because we need to exclude the elements 150 and 160. Therefore, the correct output is 260.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a monotonic decreasing stack to find nearest greater boundaries in O(1) amortized time, achieving a single linear pass.
Brute Force Approach
Check every possible sub‑matrix or transaction window and compute its consolidator value, leading to O(n^2) or higher time.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) {
break;
}
sum += nums[i];
}
for (let i = nums.length - 1; i >= 0; i--) {
if (nums[i] > K) {
sum += nums[i];
} else {
break;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > K) {
break;
}
sum += nums[i];
}
for (int i = nums.size() - 1; i >= 0; i--) {
if (nums[i] > K) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > K) {
break;
}
sum += nums[i];
}
for (int i = nums.length - 1; i >= 0; i--) {
if (nums[i] > K) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
}def solution(nums, K):
sum = 0
for i in range(len(nums)):
if nums[i] > K:
break
sum += nums[i]
for i in range(len(nums) - 1, -1, -1):
if nums[i] > K:
sum += nums[i]
else:
break
return sumfunction solution(nums, K) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) {
break;
}
sum += nums[i];
}
for (let i = nums.length - 1; i >= 0; i--) {
if (nums[i] > K) {
sum += nums[i];
} else {
break;
}
}
return sum;
}Asked in Top Tech Interviews
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