Given a m x n grid filled with integers, find the submatrix with maximum sum among all submatrices.
Example:
Input: [ [ 1, -9, -10, 1], [-1, 10, 10, 1], [ 0, 9, 9, -9], [-1, -1, -1, -1] ] Output: 38 Explanation: Submatrix [[10, 10], [9, 9]]
The core challenge of this problem is to find the submatrix within a given m x n grid that has the maximum sum. This problem is significant in various applications such as image processing, financial analysis, and more. A common pitfall is to consider only submatrices of fixed sizes or to overlook negative values that might affect the sum.
To solve this problem, we can use a dynamic programming approach combined with the concept of prefix sums. Here's a step-by-step breakdown:
A naive solution would involve checking all possible submatrices and calculating their sums. This approach is not optimal due to its high time complexity of O(m^2 * n^2 * min(m, n)).
We can optimize the solution by using prefix sums and Kadane's algorithm. The idea is to reduce the problem to a series of 1D maximum subarray sum problems.
Here is a detailed breakdown of the algorithm:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// Function to find the maximum sum submatrix
int maxSumSubmatrix(vector<vector<int>>& matrix) {
int m = matrix.size();
int n = matrix[0].size();
int maxSum = INT_MIN;
// Prefix sums for each row
vector<vector<int>> prefixSum(m, vector<int>(n + 1, 0));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
prefixSum[i][j + 1] = prefixSum[i][j] + matrix[i][j];
}
}
// Iterate over all pairs of rows
for (int top = 0; top < m; ++top) {
for (int bottom = top; bottom < m; ++bottom) {
// 1D array to store the sum of elements between the two rows for each column
vector<int> colSum(n, 0);
for (int col = 0; col < n; ++col) {
colSum[col] = prefixSum[bottom][col + 1] - prefixSum[top][col];
}
// Apply Kadane's algorithm to find the maximum subarray sum
int currentMax = colSum[0];
int currentSum = colSum[0];
for (int k = 1; k < n; ++k) {
currentSum = max(colSum[k], currentSum + colSum[k]);
currentMax = max(currentMax, currentSum);
}
// Update the maximum sum
maxSum = max(maxSum, currentMax);
}
}
return maxSum;
}
int main() {
vector<vector<int>> matrix = {
{1, -9, -10, 1},
{-1, 10, 10, 1},
{0, 9, 9, -9},
{-1, -1, -1, -1}
};
cout << "Maximum Sum Submatrix: " << maxSumSubmatrix(matrix) << endl;
return 0;
}
The time complexity of this approach is O(m^2 * n) because we iterate over all pairs of rows and apply Kadane's algorithm on each pair. The space complexity is O(m * n) due to the prefix sums array.
Potential edge cases include:
Each of these cases is handled by the algorithm, ensuring it returns the correct maximum sum.
To test the solution comprehensively, consider the following test cases:
Using a testing framework like Google Test can help automate and validate these test cases.
When approaching such problems, consider breaking them down into smaller subproblems. Use dynamic programming and prefix sums to simplify complex calculations. Practice similar problems to improve your problem-solving skills.
Understanding and solving the maximum sum submatrix problem is crucial for various applications. By using optimized algorithms like Kadane's algorithm and prefix sums, we can efficiently find the solution. Practice and exploration of similar problems will further enhance your skills.
For further reading and practice, consider the following resources: