Matrix Transaction Tracker 6 — Problem Statement & Solution Guide

GreedyHardFrequency Hash Map

TimeO(N·M log M) preprocessing + O(N·log M) per query

SpaceO(N·M) to store the sorted matrix (in‑place possible O(1) extra)

Problem Description

Given a sequence of data elements representing matrix and transaction metrics, construct an optimal algorithm to evaluate and compute the target tracker value under given operational constraints, where the tracker value is the sum of the maximum elements in each row that are less than or equal to K.

DSA Pattern Breakdown

"Matrix Transaction Tracker 6"

hard

WHY DOES IT MATTER?

The pattern exemplifies independent‑row greedy optimization, a common scenario where local optimal choices compose a global optimum. Recognizing this decoupling lets you avoid complex DP or global search and focus on per‑segment processing.

OPTIMIZATION CHALLENGE

The key insight is that sorting each row once transforms a linear “max ≤ K” search into a binary‑search problem, reducing per‑row query time from O(M) to O(log M) and enabling fast handling of multiple K queries.

REAL-WORLD CONNECTION

Think of a bank processing daily transaction limits per account: each account (row) can approve the largest transaction not exceeding the daily cap (K). The total approved amount is the sum across accounts, mirroring the matrix‑row selection problem.

When coding, first sort rows in descending order, then use std::upper_bound (or bisect_right) to locate the split point. Remember to handle rows where all elements exceed K – they contribute zero.

COMPLEXITY AT A GLANCE

⏱ Time:O(N·M log M) preprocessing + O(N·log M) per query

💾 Space:O(N·M) to store the sorted matrix (in‑place possible O(1) extra)

Core Theory — Why This Approach?

The problem reduces to a row‑wise greedy choice: for each row we must pick the greatest element that does not exceed the global threshold K, because any smaller element would only decrease the contribution of that row to the final sum while still satisfying the constraint. A naïve solution would examine every element of every row, checking the condition and keeping the maximum per row, which runs in O(N·M) time where N is the number of rows and M the number of columns. When N and M each can be as large as 10^5, the total element count can reach 10^10, making a full scan infeasible. The optimal paradigm leverages the independence of rows and the monotonic nature of the “≤ K” condition: by sorting each row once (O(M log M) per row) we can binary‑search for the rightmost element ≤ K in O(log M) time, yielding an overall O(N·(M log M + log M)) ≈ O(N·M log M) preprocessing cost, but the per‑query cost becomes O(N·log M). If the matrix is static and many queries with different K values are expected, we can pre‑sort rows and answer each query in O(N·log M). For a single K, a linear scan with early termination using a max‑heap or bucket counting (when values are bounded) can achieve O(N·M) worst‑case but with a much smaller constant factor, often passing the limits. The greedy choice is provably optimal because rows do not interact; maximizing each row independently maximizes the global sum.

Interview Questions on This Problem

Q1How would you modify the solution if each row could be used at most twice across different K queries?

Store each row sorted in descending order and maintain a pointer to the current eligible element. For each new K, move the pointer left until the element ≤ K, then record the value. If a row is used twice, keep two pointers or a usage counter to ensure the second selection skips the previously chosen element.

Q2Explain why a simple linear scan per row may still be acceptable for N = 10^5 and M = 10 when the time limit is 2 seconds.

The total number of elements is only 10^6, and a single pass over 1 million integers fits comfortably within 2 seconds in modern judges. The constant factor of a linear scan is tiny compared to sorting or binary search overhead, so the naïve O(N·M) approach is practically optimal for such small M.

Q3In a distributed system where each node holds a subset of rows, how would you compute the global tracker value efficiently?

Each node independently computes the sum of its local rows using the greedy per‑row rule (either via scan or binary search on sorted rows). Then a single reduction (e.g., MPI_Reduce or a map‑reduce combine step) aggregates the partial sums to produce the final tracker value, achieving O(local‑rows·M) work per node and O(log P) communication steps for P nodes.

Examples

Example 1

Input

[[1, 2, 3], [4, 5, 6], [7, 8, 9]] and K = 5

Output

Explanation: Step-by-step: 1. Sort each row in ascending order. 2. Iterate through each row and find the maximum element less than or equal to K. 3. Sum up the maximum elements from each row to get the tracker value.

Example 2

Input

[[10, 20, 30], [40, 50, 60], [70, 80, 90]] and K = 50

Output

Constraints

1 <= N <= 10^5
-10^4 <= metrics[i] <= 10^4
1 <= K <= N

Optimal Approach & Strategy

Sort each row descending once, then binary‑search for the rightmost element ≤ K in each row and accumulate the results.

Brute Force Approach

Iterate every element, keep a per‑row maximum that satisfies ≤ K, then sum those maxima.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N·M log M) preprocessing + O(N·log M) per query

function solution(matrix, K) {
   let trackerValue = 0;
   for (let row of matrix) {
       row.sort((a, b) => a - b);
       let maxElement = row[row.length - 1];
       if (maxElement <= K) trackerValue += maxElement;
   }
   return trackerValue;
}

class Solution {
   public:
   int solution(vector<vector<int>>& matrix, int K) {
       int trackerValue = 0;
       for (vector<int>& row : matrix) {
           sort(row.begin(), row.end());
           int maxElement = row.back();
           if (maxElement <= K) trackerValue += maxElement;
       }
       return trackerValue;
   }
}

class Solution {
   public int solution(int[][] matrix, int K) {
       int trackerValue = 0;
       for (int[] row : matrix) {
           Arrays.sort(row);
           int maxElement = row[row.length - 1];
           if (maxElement <= K) trackerValue += maxElement;
       }
       return trackerValue;
   }
}

def solution(matrix, K):
   tracker_value = 0
   for row in matrix:
       row.sort()
       max_element = row[-1]
       if max_element <= K: tracker_value += max_element
   return tracker_value

function solution(matrix, K) {
   let trackerValue = 0;
   for (let row of matrix) {
       row.sort((a, b) => a - b);
       let maxElement = row[row.length - 1];
       if (maxElement <= K) trackerValue += maxElement;
   }
   return trackerValue;
}

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Matrix Transaction Tracker 6 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Greedy

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Matrix Transaction Tracker 6 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor