Maximum Sum Submatrix III in C++ (Time Complexity: O(m^2 * n^2))

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]]

Understanding the Problem

The core challenge of this problem is to identify 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 any domain where analyzing sub-regions of a matrix is crucial. A common pitfall is to assume that the submatrix must be contiguous in both rows and columns, which is indeed the case here.

Approach

To solve this problem, we can use a dynamic programming approach combined with the concept of prefix sums. The naive solution would involve checking all possible submatrices, which is computationally expensive with a time complexity of O(m^3 * n^3). Instead, we can optimize this by reducing the problem to a series of 1D maximum subarray problems (Kadane's algorithm).

Optimized Solution

1. **Prefix Sum Calculation**: First, we calculate the prefix sums for each row to quickly compute the sum of any submatrix. 2. **Kadane's Algorithm**: For each pair of rows, we use Kadane's algorithm to find the maximum sum subarray for the columns between these rows.

Algorithm

1. Compute the prefix sums for each row. 2. Iterate over all pairs of rows. 3. For each pair of rows, use Kadane's algorithm to find the maximum sum subarray for the columns between these rows. 4. Keep track of the maximum sum found.

Code Implementation


#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) {
            // Use Kadane's algorithm to find the maximum sum subarray
            int currentSum = 0;
            int maxSubarraySum = INT_MIN;
            for (int col = 0; col < n; ++col) {
                int colSum = prefixSum[bottom][col + 1] - prefixSum[top][col + 1];
                currentSum = max(colSum, currentSum + colSum);
                maxSubarraySum = max(maxSubarraySum, currentSum);
            }
            maxSum = max(maxSum, maxSubarraySum);
        }
    }

    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;
}

Complexity Analysis

The time complexity of this approach is O(m^2 * n) because we iterate over all pairs of rows (O(m^2)) and for each pair, we apply Kadane's algorithm on the columns (O(n)). The space complexity is O(m * n) due to the prefix sum matrix.

Edge Cases

Potential edge cases include:

All elements are negative: The algorithm should still find the submatrix with the least negative sum.
Single row or single column matrices: The algorithm should handle these gracefully.
Empty matrix: The algorithm should return 0 or an appropriate value indicating no submatrix exists.

Testing

To test the solution comprehensively, consider the following test cases:

Matrix with all positive numbers.
Matrix with all negative numbers.
Matrix with mixed positive and negative numbers.
Single row and single column matrices.
Empty matrix.

Thinking and Problem-Solving Tips

When approaching such problems:

Break down the problem into smaller subproblems.
Use dynamic programming to store intermediate results and avoid redundant calculations.
Think about how to reduce the problem to a simpler, well-known problem (e.g., reducing 2D submatrix sum to 1D subarray sum).
Practice similar problems to strengthen your understanding and problem-solving skills.

Conclusion

Understanding and solving the maximum sum submatrix problem is crucial for various applications. By breaking down the problem and using efficient algorithms like Kadane's, we can significantly optimize our solution. Practice and familiarity with dynamic programming and prefix sums are key to mastering such problems.

Additional Resources

For further reading and practice: