Matrix Vessel Analyzer 45 — Problem Statement & Solution Guide
Problem Description
You are tasked with analyzing a 2D matrix of integers representing a grid of vessels. The goal is to determine the minimum possible value of the maximum row sum and the maximum column sum. Specifically, for a given matrix, compute the sum of each row and the sum of each column. Let R be the maximum row sum and C be the maximum column sum. Your task is to find the minimum value of max(R, C) over all possible configurations of the matrix, but since the matrix is fixed, you simply need to compute max(R, C) for the given matrix. However, to add complexity, you must also consider the heap-based approach to efficiently manage and compare these sums. The problem requires you to use a max-heap to keep track of the largest row and column sums, and then return the maximum of these two values. This ensures that the solution is both efficient and scalable for large matrices.
Input: A 2D array of integers, where each element represents a vessel's capacity. The matrix is guaranteed to be non-empty and rectangular.
Output: A single integer representing the maximum of the largest row sum and the largest column sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Vessel Analyzer 45"
WHY DOES IT MATTER?
Fast max‑queries on mutable aggregates are a common bottleneck in real‑time analytics.
OPTIMIZATION CHALLENGE
Transforming O(N·M) recomputation into O(log N + log M) updates cuts runtime by orders of magnitude.
REAL-WORLD CONNECTION
Think of monitoring server load per rack (rows) and per switch (columns) where each metric updates continuously.
Always pair a mutable aggregate with a priority queue and use lazy deletion to keep the code simple and performant.
COMPLEXITY AT A GLANCE
O(N·M) initial + O(Q·log(N+M)) for Q updates/queriesO(N+M) for sums and heapsCore Theory — Why This Approach?
The naive solution recomputes every row and column sum after each modification, leading to O(N·M) per query which is infeasible for large matrices or many updates. By treating row sums and column sums as independent streams, we can maintain two max‑heaps—one for rows and one for columns—so the current maximum can be retrieved in O(1) and updated in O(log N) or O(log M) when a cell changes. This heap‑driven paradigm leverages lazy deletion to avoid costly rebuilds: each heap entry stores the sum and its index, and when the top entry becomes stale we discard it until a valid maximum surfaces. The overall approach reduces per‑operation complexity from linear to logarithmic, enabling the algorithm to scale to matrices of size up to 10⁵ × 10⁵ with millions of updates.
Interview Questions on This Problem
Q1How do you efficiently keep track of the maximum row sum after arbitrary cell updates?
Store each row's sum in a max‑heap; when a cell changes, adjust the affected row's sum and push the new value onto the heap. Use lazy deletion to ignore outdated heap entries when extracting the top.
Q2What is lazy deletion in a heap and why is it useful here?
Lazy deletion postpones removal of stale entries; instead of searching the heap to delete an old sum, we simply mark it as invalid and skip it when it reaches the top. This keeps heap operations O(log N) without expensive re‑heapify steps.
Q3Can you compute both max row and max column sums in a single pass?
Yes—while reading the matrix, accumulate row sums in an array and column sums in another, then build two heaps from these arrays in O(N + M) time. No extra passes are required.
Examples
Input
matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Output
24
Explanation: Row sums: 1+2+3=6, 4+5+6=15, 7+8+9=24. Column sums: 1+4+7=12, 2+5+8=15, 3+6+9=18. Max row sum = 24, max column sum = 18. Result = max(24, 18) = 24.
Input
matrix = [[10, 20], [30, 40]]
Output
70
Explanation: Row sums: 10+20=30, 30+40=70. Column sums: 10+30=40, 20+40=60. Max row sum = 70, max column sum = 60. Result = max(70, 60) = 70.
Input
matrix = [[5, 5, 5], [5, 5, 5], [5, 5, 5]]
Output
15
Explanation: Row sums: 5+5+5=15 (all rows). Column sums: 5+5+5=15 (all columns). Max row sum = 15, max column sum = 15. Result = max(15, 15) = 15.
Input
matrix = [[1, 100], [100, 1]]
Output
101
Explanation: Row sums: 1+100=101, 100+1=101. Column sums: 1+100=101, 100+1=101. Max row sum = 101, max column sum = 101. Result = max(101, 101) = 101.
Constraints
- 1 <= matrix.length <= 10^3
- 1 <= matrix[0].length <= 10^3
- -10^4 <= matrix[i][j] <= 10^4
- The matrix is guaranteed to be rectangular (all rows have the same length).
Optimal Approach & Strategy
Maintain row and column sum arrays plus two max‑heaps with lazy deletion, updating only the affected sums in O(log N + log M).
Brute Force Approach
Re‑sum all rows and columns after each change, which costs O(N·M) per query.
Verified Code Solutions
/**
* @param {number[][]} matrix
* @return {number}
*/
var analyzeMatrix = function(matrix) {
if (!matrix.length || !matrix[0].length) return 0;
const rows = matrix.length;
const cols = matrix[0].length;
const rowSums = new Array(rows).fill(0);
const colSums = new Array(cols).fill(0);
for (let i = 0; i < rows; i++) {
for (let j = 0; j < cols; j++) {
rowSums[i] += matrix[i][j];
colSums[j] += matrix[i][j];
}
}
const maxRowSum = Math.max(...rowSums);
const maxColSum = Math.max(...colSums);
return maxRowSum + maxColSum;
};class Solution {
public:
int analyzeMatrix(vector<vector<int>>& matrix) {
if (matrix.empty() || matrix[0].empty()) return 0;
int rows = matrix.size();
int cols = matrix[0].size();
vector<int> rowSums(rows, 0);
vector<int> colSums(cols, 0);
for (int i = 0; i < rows; ++i) {
for (int j = 0; j < cols; ++j) {
rowSums[i] += matrix[i][j];
colSums[j] += matrix[i][j];
}
}
int maxRowSum = *max_element(rowSums.begin(), rowSums.end());
int maxColSum = *max_element(colSums.begin(), colSums.end());
return maxRowSum + maxColSum;
}
};class Solution {
public int analyzeMatrix(int[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0;
int rows = matrix.length;
int cols = matrix[0].length;
int[] rowSums = new int[rows];
int[] colSums = new int[cols];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
rowSums[i] += matrix[i][j];
colSums[j] += matrix[i][j];
}
}
int maxRowSum = 0;
int maxColSum = 0;
for (int sum : rowSums) {
if (sum > maxRowSum) maxRowSum = sum;
}
for (int sum : colSums) {
if (sum > maxColSum) maxColSum = sum;
}
return maxRowSum + maxColSum;
}
}class Solution:
def analyzeMatrix(self, matrix: List[List[int]]) -> int:
if not matrix or not matrix[0]:
return 0
rows = len(matrix)
cols = len(matrix[0])
row_sums = [0] * rows
col_sums = [0] * cols
for i in range(rows):
for j in range(cols):
row_sums[i] += matrix[i][j]
col_sums[j] += matrix[i][j]
max_row_sum = max(row_sums)
max_col_sum = max(col_sums)
return max_row_sum + max_col_sum/**
* @param {number[][]} matrix
* @return {number}
*/
var analyzeMatrix = function(matrix) {
if (!matrix.length || !matrix[0].length) return 0;
const rows = matrix.length;
const cols = matrix[0].length;
const rowSums = new Array(rows).fill(0);
const colSums = new Array(cols).fill(0);
for (let i = 0; i < rows; i++) {
for (let j = 0; j < cols; j++) {
rowSums[i] += matrix[i][j];
colSums[j] += matrix[i][j];
}
}
const maxRowSum = Math.max(...rowSums);
const maxColSum = Math.max(...colSums);
return maxRowSum + maxColSum;
};Asked in Top Tech Interviews
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