Matrix Transaction Architect 20 — 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 architect value under given operational constraints. The algorithm should handle the case when K is greater than the array length correctly, handle the case when the array contains non-numeric values or when K is a non-numeric value, and add values greater than K to the sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Transaction Architect 20"
WHY DOES IT MATTER?
Binary search on sorted matrices reduces search time from linear to logarithmic, a critical performance gain for big‑data applications.
OPTIMIZATION CHALLENGE
The key is mapping a 1‑D index to 2‑D coordinates without extra storage, cutting the search complexity by a factor of log (m·n).
REAL-WORLD CONNECTION
Databases use similar index‑flattening techniques to locate rows in multi‑dimensional B‑trees efficiently.
Always validate input dimensions and data types upfront; a single guard clause saves hours of debugging later.
COMPLEXITY AT A GLANCE
O(log(m·n))O(1)Core Theory — Why This Approach?
Binary search exploits the monotonic ordering of data to halve the search space at each step, yielding logarithmic time complexity. When a matrix is row‑wise and column‑wise sorted, it can be treated as a flattened sorted array, allowing a classic O(log (m·n)) binary search across the virtual index space, which is far superior to linear scans on large datasets. Naïve approaches—such as iterating every element or performing a separate binary search per row—degenerate to O(m·n) or O(m·log n) time, quickly exhausting time limits for matrices with millions of entries. The optimal paradigm leverages the matrix’s global ordering, maps a 1‑D index to 2‑D coordinates via division and modulo, and applies a single binary search loop that respects the given operational constraints, including handling out‑of‑bounds K values and non‑numeric entries through early validation.
Interview Questions on This Problem
Q1How can you perform binary search on a 2‑D matrix without extra space?
Treat the matrix as a virtual sorted array of size m·n and compute row = mid / n, col = mid % n for each mid index. This preserves O(1) space and O(log (m·n)) time.
Q2Why does a row‑wise binary search fail to achieve optimal complexity on a fully sorted matrix?
Each row search is O(log n) and repeated for m rows, resulting in O(m·log n) which is slower than a single O(log (m·n)) search that exploits the global ordering.
Q3What edge cases must be guarded against when K exceeds the total number of elements?
Validate K against m·n before the search; if K > m·n, return an error or sentinel value. This prevents index overflow and undefined behavior.
Examples
Input
[1, 2, 3, 4, 5, 10]
Output
25
Explanation: Step-by-step: Given the array [1, 2, 3, 4, 5, 10] and K = 5, we iterate through the array from left to right. We add 6 (which is greater than K) to the sum, giving us a final sum of 25.
Input
[10, 20, 30, 40, 50]
Output
0
Explanation: Step-by-step: Given the array [10, 20, 30, 40, 50] and K = 60, we iterate through the array from left to right. Since all elements are less than K, we do not add any elements to the sum, giving us a final sum of 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Apply a single binary search on the virtual 1‑D view of the matrix, converting indices on the fly for O(log (m·n)) time and O(1) space.
Brute Force Approach
Iterate every cell sequentially until the target is found, which is O(m·n) time.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (typeof nums[i] !== 'number' || typeof k !== 'number') {
throw new Error('Input contains non-numeric values');
}
if (nums[i] > k) {
sum += nums[i];
}
}
if (k < Math.min(...nums)) {
throw new Error('K is less than the smallest element in the array');
}
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] >= 0 && nums[i] <= INT_MAX) || !(k >= 0 && k <= INT_MAX)) {
throw std::invalid_argument('Input contains non-numeric values');
}
if (nums[i] > k) {
sum += nums[i];
}
}
if (k < *min_element(nums.begin(), nums.end())) {
throw std::invalid_argument('K is less than the smallest element in the array');
}
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] instanceof Integer) || !(k instanceof Integer)) {
throw new IllegalArgumentException('Input contains non-numeric values');
}
if (nums[i] > k) {
sum += nums[i];
}
}
if (k < Arrays.stream(nums).min().getAsInt()) {
throw new IllegalArgumentException('K is less than the smallest element in the array');
}
return sum;
}
}def solution(nums, k):
sum = 0
for i in range(len(nums)):
if not isinstance(nums[i], (int, float)) or not isinstance(k, (int, float)):
raise ValueError('Input contains non-numeric values')
if nums[i] > k:
sum += nums[i]
if k < min(nums):
raise ValueError('K is less than the smallest element in the array')
return sumfunction solution(nums, k) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (typeof nums[i] !== 'number' || typeof k !== 'number') {
throw new Error('Input contains non-numeric values');
}
if (nums[i] > k) {
sum += nums[i];
}
}
if (k < Math.min(...nums)) {
throw new Error('K is less than the smallest element in the array');
}
return sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
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