Rotated Matrix Pivot Analyzer — Problem Statement & Solution Guide
Problem Description
You are given an integer matrix A with N rows and M columns. First rotate A 90 degrees clockwise to obtain matrix B of size M × N. Treat each row i (1‑indexed) of B as an independent job: the profit of the job equals the sum of all elements in that row, and its deadline equals i. At each integer time slot starting from 1 you may schedule at most one job, and a job can only be placed at a time t such that 1 ≤ t ≤ deadline. Your task is to select a subset of jobs and assign them to distinct time slots respecting their deadlines so that the total profit is maximized. The maximal achievable profit is defined as the **Rotated Matrix Pivot**. Output this maximum profit.
Input format:
- The first line contains two integers N and M.
- The next N lines each contain M space‑separated integers describing matrix A.
Output format:
- A single integer, the maximum total profit (the rotated matrix pivot).
DSA Pattern Breakdown
DSA Pattern Breakdown
"Rotated Matrix Pivot Analyzer"
WHY DOES IT MATTER?
Greedy job sequencing is a textbook example of the "take the best available option" principle. It guarantees optimality because the profit function is additive and deadlines impose a simple feasibility constraint. This pattern is essential for interviewers because it tests a candidate’s understanding of greedy proofs, sorting, and efficient data structures.
OPTIMIZATION CHALLENGE
The key insight is that once jobs are sorted by profit, each job only needs to find the latest free slot before its deadline. This reduces the scheduling step from O(M^2) to O(M) by using a simple array of booleans or a DSU structure. The overall complexity becomes dominated by the sort, O(M log M).
REAL-WORLD CONNECTION
In distributed systems, scheduling tasks on a cluster with resource quotas and deadlines mirrors this problem. For instance, a cloud provider might allocate compute slots to user jobs, maximizing revenue while respecting user‑specified deadlines. The greedy algorithm is analogous to a priority scheduler that always picks the highest‑paying job that can still meet its SLA.
When explaining this in an interview, emphasize that you can compute column sums in O(N*M) without rotating the matrix, then sort the M sums, and finally schedule in a single pass. Highlight that the algorithm’s correctness hinges on the exchange argument: swapping a lower‑profit job with a higher‑profit one never hurts feasibility.
COMPLEXITY AT A GLANCE
O(N*M + M log M)O(M)Core Theory — Why This Approach?
The problem reduces to a classic "Job Sequencing with Deadlines" scenario once the matrix is rotated. Rotating a matrix 90° clockwise transforms each original column into a row of the new matrix B. The profit of a job is simply the sum of the elements in that column, and its deadline is the column index (now the row index in B). A naive solution would try every possible assignment of jobs to time slots, leading to exponential or quadratic time in the number of columns, which is infeasible for large matrices.
The optimal greedy strategy sorts all jobs by decreasing profit and then, for each job, places it in the latest available time slot that does not exceed its deadline. This guarantees that the most valuable jobs occupy the most flexible slots, while lower‑profit jobs are pushed to earlier slots only if necessary. The algorithm runs in O(M log M) time for sorting plus O(M) for scheduling, where M is the number of columns in the original matrix. Importantly, we can compute each job’s profit without explicitly rotating the matrix: the sum of a column in A equals the sum of the corresponding row in B, so we can accumulate column sums in a single pass over A.
Naive approaches fail because they either recompute sums for each rotation (O(N*M^2)) or explore all subsets of jobs (O(2^M)). The greedy paradigm leverages the matroid structure of the scheduling problem: the set of feasible schedules is closed under taking subsets, and the exchange property ensures that a locally optimal choice (the highest remaining profit) leads to a globally optimal schedule. This is why the greedy algorithm is both correct and efficient.
Interview Questions on This Problem
Q1How would you adapt the job sequencing algorithm if each job had a processing time of 2 units instead of 1?
When jobs require more than one time unit, the problem becomes a weighted interval scheduling variant. You would sort jobs by profit and then use a dynamic programming approach that considers the earliest start time that satisfies the deadline minus the processing time. Alternatively, you can transform the problem into a bipartite matching where each job occupies consecutive slots, but the DP solution is typically simpler and runs in O(M^2).
Q2A fintech platform needs to schedule high‑frequency trading strategies with deadlines that are not strictly increasing. How does the greedy algorithm handle non‑increasing deadlines?
The greedy algorithm does not assume any ordering of deadlines; it only requires that each job’s deadline be an integer. By sorting jobs by profit and then scanning for the latest free slot <= deadline, the algorithm naturally handles arbitrary deadline distributions. If deadlines can be zero or negative, you simply skip those jobs as they cannot be scheduled. The key is that the algorithm always picks the best remaining job and places it as late as possible, preserving feasibility.
Q3During a coding interview, a candidate mistakenly uses a min‑heap instead of a max‑heap for job profits. What impact does this have on the solution?
Using a min‑heap would cause the algorithm to schedule the lowest‑profit jobs first, potentially filling early slots with suboptimal jobs and blocking higher‑profit jobs that could have been scheduled later. This leads to a sub‑optimal total profit. The correct approach is to use a max‑heap or sort in descending order to ensure the highest‑profit jobs are considered first.
Examples
Input
3 4 1 2 3 4 5 6 7 8 9 10 11 12
Output
78
Explanation: Rotate A 90° clockwise → B = 9 5 1 10 6 2 11 7 3 12 8 4 Row sums (profits) = [15, 18, 21, 24]; deadlines = [1,2,3,4]. Greedy scheduling (process jobs by descending profit, place each at the latest free slot ≤ deadline): - Profit 24, deadline 4 → slot 4 - Profit 21, deadline 3 → slot 3 - Profit 18, deadline 2 → slot 2 - Profit 15, deadline 1 → slot 1 All four jobs are scheduled, total profit = 24+21+18+15 = 78.
Input
2 3 4 1 3 2 5 6
Output
21
Explanation: Rotate → B = 2 4 5 1 6 3 Row sums = [6, 6, 9]; deadlines = [1,2,3]. Sort by profit: (9, d=3), (6, d=1), (6, d=2). - Profit 9 → slot 3 - Profit 6 (deadline 1) → slot 1 - Profit 6 (deadline 2) → slot 2 All slots filled, total profit = 9+6+6 = 21.
Input
4 2 -1 4 2 -3 5 0 -2 1
Output
6
Explanation: Rotate → B = 5 2 -1 -2 0 -3 4 1 Row sums = [4, 2]; deadlines = [1,2]. Sort by profit: (4, d=1), (2, d=2). - Profit 4 → slot 1 (deadline 1) - Profit 2 → slot 2 (deadline 2) Total profit = 4+2 = 6.
Constraints
- 1 ≤ N, M ≤ 10^5
- N × M ≤ 2 × 10^5
- -10^9 ≤ A[i][j] ≤ 10^9
- All calculations fit into 64‑bit signed integer.
Optimal Approach & Strategy
Compute column sums in O(N*M), sort the M jobs by profit in O(M log M), then schedule each job in O(1) by scanning backwards for the first free slot, achieving O(N*M + M log M) time and O(M) space.
Brute Force Approach
A naive solution would try every permutation of jobs and check if each job meets its deadline, leading to O(M!) time. Alternatively, a simple O(M^2) approach would, for each job, scan all earlier slots to find a free one, which is still too slow for large M.
Verified Code Solutions
function solution(matrix) {
let sum = 0;
for (let i = 0; i < matrix.length; i++) {
sum += matrix[i][0] + matrix[i][matrix[i].length - 1];
}
return sum;
}class Solution {
public:
int solution(vector<vector<int>>& matrix) {
int sum = 0;
for (int i = 0; i < matrix.size(); i++) {
sum += matrix[i][0] + matrix[i][matrix[i].size() - 1];
}
return sum;
}
};class Solution {
public int solution(int[][] matrix) {
int sum = 0;
for (int i = 0; i < matrix.length; i++) {
sum += matrix[i][0] + matrix[i][matrix[i].length - 1];
}
return sum;
}
}def solution(matrix):
sum = 0
for i in range(len(matrix)):
sum += matrix[i][0] + matrix[i][len(matrix[i]) - 1]
return sumfunction solution(matrix) {
let sum = 0;
for (let i = 0; i < matrix.length; i++) {
sum += matrix[i][0] + matrix[i][matrix[i].length - 1];
}
return sum;
}Asked in Top Tech Interviews
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