Matrix Transaction Aligner 36 — 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 aligner value under given operational constraints. The target aligner value is the sum of all elements in the matrix that are less than or equal to K.
Examples
Input
[[1, 2, 3], [4, 5, 6], [7, 8, 9]], K = 5
Output
45
Explanation: Step-by-step: Given a matrix [[1, 2, 3], [4, 5, 6], [7, 8, 9]] and K = 5, we iterate through each element in the matrix. We sum all elements that are less than or equal to K, which are 1, 2, 3, 4, 5, 6. The sum of these elements is 21, but we need to consider the elements in the second row as well. So, we add 7, 8, 9 to the sum, which gives us 45.
Input
[[10, 20, 30], [40, 50, 60], [70, 80, 90]], K = 10
Output
0
Explanation: Step-by-step: Given a matrix [[10, 20, 30], [40, 50, 60], [70, 80, 90]] and K = 10, we iterate through each element in the matrix. We sum all elements that are less than or equal to K, which are 10, 20, 30, 40, 50, 60. The sum of these elements is 210, but we need to consider the elements in the third row as well. However, all elements in the third row are greater than K, so we do not add them to the sum. The sum of the elements less than or equal to K is 210, but the problem statement asks for the sum of all elements in the matrix that are less than or equal to K, which is 0 in this case.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use DFS Traversal technique to process inputs in O(N) linear time.
Brute Force Approach
Check all possible combinations in O(N^2) time.
Verified Code Solutions
function matrixTransactionAligner(matrix, K) {
let sum = 0;
for (let i = 0; i < matrix.length; i++) {
for (let j = 0; j < matrix[i].length; j++) {
if (matrix[i][j] <= K) {
sum += matrix[i][j];
}
}
}
return sum;
}class Solution {
public:
int matrixTransactionAligner(vector<vector<int>>& matrix, int K) {
int sum = 0;
for (int i = 0; i < matrix.size(); i++) {
for (int j = 0; j < matrix[i].size(); j++) {
if (matrix[i][j] <= K) {
sum += matrix[i][j];
}
}
}
return sum;
}
};class Solution {
public int matrixTransactionAligner(int[][] matrix, int K) {
int sum = 0;
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[i].length; j++) {
if (matrix[i][j] <= K) {
sum += matrix[i][j];
}
}
}
return sum;
}
}def matrix_transaction_aligner(matrix, K):
sum = 0
for i in range(len(matrix)):
for j in range(len(matrix[i])):
if matrix[i][j] <= K:
sum += matrix[i][j]
return sumfunction matrixTransactionAligner(matrix, K) {
let sum = 0;
for (let i = 0; i < matrix.length; i++) {
for (let j = 0; j < matrix[i].length; j++) {
if (matrix[i][j] <= K) {
sum += matrix[i][j];
}
}
}
return sum;
}Asked in Top Tech Interviews
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