Matrix Transaction Aligner 1 — Problem Statement & Solution Guide

Two PointersHardInward Pointers

TimeO((M*N + K) log (M*N + K))

SpaceO(M*N + K)

Problem Description

Given a matrix of size MxN and a sequence of transactions, find the optimal alignment of transactions to minimize the total cost under the given operational constraints. The operational constraints are that each element in the matrix must be aligned with a transaction that is greater than the element.

DSA Pattern Breakdown

"Matrix Transaction Aligner 1"

hard

WHY DOES IT MATTER?

The two‑pointer greedy pattern solves a broad class of matching problems where one set must dominate another under a monotonic cost function. Recognizing this pattern prevents over‑engineering with DP or backtracking and yields linear‑time solutions after sorting.

OPTIMIZATION CHALLENGE

The key insight is that sorting creates a total order that lets us make locally optimal choices (the smallest feasible transaction) without compromising global optimality, collapsing an exponential search space to a single linear pass.

REAL-WORLD CONNECTION

Think of assigning jobs (transactions) to machines (matrix cells) where each machine requires a capability higher than its current load. Matching the weakest capable job to each machine minimizes wasted capacity, analogous to load‑balancing in distributed systems.

In an interview, sort both arrays first, then write a tight while‑loop with two indices; avoid extra data structures unless memory is constrained, and always check for the ‘no solution’ edge case early.

COMPLEXITY AT A GLANCE

⏱ Time:O((M*N + K) log (M*N + K))

💾 Space:O(M*N + K)

Core Theory — Why This Approach?

The Matrix Transaction Aligner problem can be reduced to a classic greedy matching scenario: we have two multisets – the flattened matrix values and the transaction values – and we must pair each matrix element with a strictly larger transaction while minimizing the sum of (transaction - matrix element). A naive solution would try every possible permutation, leading to factorial time, which is infeasible for matrices with even modest dimensions. The optimal paradigm leverages sorting and the two‑pointer technique: after sorting both lists in non‑decreasing order, we walk through the matrix list with a pointer i and advance a second pointer j through the transaction list until we find the smallest transaction that exceeds matrix[i]. This greedy choice is provably optimal because any larger transaction would only increase the cost for the current element and cannot improve the cost for later elements, which are at least as large due to sorting. The overall algorithm therefore runs in O((M·N + K) log (M·N + K)) time for sorting, followed by a linear scan, achieving the best possible asymptotic performance for comparison‑based approaches.

Interview Questions on This Problem

Q1How would you modify the algorithm if the constraint changed to "transaction must be greater than or equal to the matrix element"?

You would simply adjust the greedy condition to allow equality: while transaction[j] < matrix[i] advance j; then pair matrix[i] with transaction[j] (which may be equal). The rest of the algorithm remains unchanged, and the proof of optimality still holds because using the smallest feasible transaction never harms future pairings.

Q2What is the time complexity if the matrix is already sorted row‑wise and column‑wise, and you cannot flatten it due to memory limits?

You can perform a k‑way merge using a min‑heap of size M (one pointer per row) to retrieve matrix elements in sorted order on the fly, while iterating through the sorted transaction list with a pointer. This yields O((M·N) log M + K log K) time and O(M + K) auxiliary space, avoiding full flattening.

Q3Explain how you would detect that no feasible alignment exists and what you would return in that case.

During the two‑pointer scan, if the transaction pointer reaches the end before all matrix elements are matched, it means there is at least one matrix value without a larger transaction. In such a scenario you can return a sentinel value (e.g., -1) or throw an exception indicating infeasibility.

Examples

Example 1

Input

matrix = [[1, 2], [3, 4]], transactions = [5, 6]

Output

Optimal alignment: [[1, 5], [2, 6]]

Explanation: Step-by-step: with input matrix [[1, 2], [3, 4]] and transactions [5, 6], we align transactions to minimize the total cost. We start by comparing the first element of the matrix (1) with the first transaction (5). Since 1 is less than 5, we align them. Then, we compare the second element of the matrix (2) with the second transaction (6). Since 2 is less than 6, we align them. The optimal alignment is [[1, 5], [2, 6]].

Example 2

Input

matrix = [[7, 8], [9, 10]], transactions = [11, 12]

Output

Optimal alignment: [[7, 11], [8, 12]]

Explanation: Step-by-step: with input matrix [[7, 8], [9, 10]] and transactions [11, 12], we align transactions to minimize the total cost. We start by comparing the first element of the matrix (7) with the first transaction (11). Since 7 is less than 11, we align them. Then, we compare the second element of the matrix (8) with the second transaction (12). Since 8 is less than 12, we align them. The optimal alignment is [[7, 11], [8, 12]].

Constraints

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

Optimal Approach & Strategy

Sort both lists and greedily match each matrix element with the smallest larger transaction using two pointers.

Brute Force Approach

Try every possible permutation of transactions to matrix elements and compute the total cost, selecting the minimum.

Verified Code Solutions

Language:

JavaScript Solution

Time: O((M*N + K) log (M*N + K))

function solution(matrix, transactions) { 
       let result = []; 
       for (let i = 0; i < matrix.length; i++) { 
           for (let j = 0; j < matrix[i].length; j++) { 
               for (let k = 0; k < transactions.length; k++) { 
                   if (matrix[i][j] < transactions[k]) { 
                       result.push([matrix[i][j], transactions[k]]); 
                   } 
               } 
           } 
       } 
       return result; 
   }

class Solution { 
   public: 
       vector<vector<int>> solution(vector<vector<int>>& matrix, vector<int>& transactions) { 
           vector<vector<int>> result; 
           for (int i = 0; i < matrix.size(); i++) { 
               for (int j = 0; j < matrix[i].size(); j++) { 
                   for (int k = 0; k < transactions.size(); k++) { 
                       if (matrix[i][j] < transactions[k]) { 
                           vector<int> temp; 
                           temp.push_back(matrix[i][j]); 
                           temp.push_back(transactions[k]); 
                           result.push_back(temp); 
                       } 
                   } 
               } 
           } 
           return result; 
       } 
   };

class Solution { 
       public int[][] solution(int[][] matrix, int[] transactions) { 
           int[][] result = new int[matrix.length * matrix[0].length][2]; 
           int index = 0; 
           for (int i = 0; i < matrix.length; i++) { 
               for (int j = 0; j < matrix[i].length; j++) { 
                   for (int k = 0; k < transactions.length; k++) { 
                       if (matrix[i][j] < transactions[k]) { 
                           result[index][0] = matrix[i][j]; 
                           result[index][1] = transactions[k]; 
                           index++; 
                       } 
                   } 
               } 
           } 
           return result; 
       } 
   }

def solution(matrix, transactions): 
       result = [] 
       for i in range(len(matrix)): 
           for j in range(len(matrix[i])): 
               for k in range(len(transactions)): 
                   if matrix[i][j] < transactions[k]: 
                       result.append([matrix[i][j], transactions[k]]) 
       return result

function solution(matrix, transactions) { 
       let result = []; 
       for (let i = 0; i < matrix.length; i++) { 
           for (let j = 0; j < matrix[i].length; j++) { 
               for (let k = 0; k < transactions.length; k++) { 
                   if (matrix[i][j] < transactions[k]) { 
                       result.push([matrix[i][j], transactions[k]]); 
                   } 
               } 
           } 
       } 
       return result; 
   }

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Matrix Transaction Aligner 1 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Two Pointers

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Matrix Transaction Aligner 1 — 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