Problem

Given an m x n integer matrix, find the sub-rectangle (contiguous rows and columns) with the maximum possible sum and return that sum. Optionally return the rectangle coordinates (top, left) and (bottom, right).

Examples

Example 1

$$ \begin{array}{|c|c|c|c|} \hline 1 & {{-2}} & {{0}} & {{-15}} \\ \hline 2 & {\color{green}{3}} & {\color{green}{-1}} & {\color{green}{4}} \\ \hline -1 & \color{green}2 & \color{green}5 & \color{green}-3 \\ \hline \end{array} $$

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Input:
mat = [
 [1, -2, 0, -15],
 [2,  3, -1, 4],
 [-1, 2,  5, -3]
]
Output: 10
Explanation: The maximum-sum rectangle is the 2x3 sub-rectangle covering rows 1..2 and columns 1..3, shown in green in diagram.
Similar Problems

Solution

We show two standard approaches: a brute-force enumeration and an efficient reduction to 1D Kadane per column-pair.

Method 1 — Brute-force enumeration

Intuition

Try every rectangle and compute its sum. Straightforward but expensive.

Approach

  • Enumerate top and bottom row indices.
  • For each top..bottom range enumerate left and right column indices.
  • Compute rectangle sum (use 2D prefix sums to compute each rectangle in O(1)).
  • Track the maximum sum and coordinates.

Code

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// Brute-force with optional 2D prefix sums
class Solution {
 public:
    long long maxSumSubmatrix(const std::vector<std::vector<int>>& A) {
        int rows = A.size();
        if (rows == 0) return 0;
        int cols = A[0].size();
        // build prefix sums (PS[i+1][j+1] = sum of A[0..i][0..j])
        std::vector<std::vector<long long>> PS(rows+1, std::vector<long long>(cols+1, 0));
        for (int i = 0; i < rows; ++i)
            for (int j = 0; j < cols; ++j)
                PS[i+1][j+1] = PS[i+1][j] + PS[i][j+1] - PS[i][j] + A[i][j];

        long long best = LLONG_MIN;
        for (int top = 0; top < rows; ++top) {
            for (int bottom = top; bottom < rows; ++bottom) {
                for (int left = 0; left < cols; ++left) {
                    for (int right = left; right < cols; ++right) {
                        long long sum = PS[bottom+1][right+1] - PS[top][right+1] - PS[bottom+1][left] + PS[top][left];
                        best = std::max(best, sum);
                    }
                }
            }
        }
        return best;
    }
};

Complexity

  • ⏰ Time complexity: O(rows^2 * cols^2) — four nested loops over rectangle boundaries; with prefix sums each rectangle sum is O(1).
  • 🧺 Space complexity: O(rows * cols) for the prefix-sum table.

Method 2 — Fix columns and run Kadane on rows (recommended)

Intuition

Fix two column boundaries left and right. Collapse the columns between them into a 1D array tmp[row] holding the row-wise sums between left and right. The maximum-sum rectangle that spans those columns is the maximum subarray of tmp (Kadane). Iterate all column pairs.

Approach

  1. For each left from 0..cols-1:
    • Initialize tmp as a zero array of length rows.
  2. For each right from left..cols-1:
    • For each row r, do tmp[r] += A[r][right].
    • Run Kadane on tmp to get the maximum contiguous row-range sum and (optionally) top/bottom indices.
    • Update global maximum and store coordinates if needed.

Edge cases: empty matrix, single row/column, all-negative elements (Kadane handles this by returning the maximum element).

Code

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class Solution {
 public:
    long long maxSumSubmatrix(const std::vector<std::vector<int>>& A) {
        int rows = A.size();
        if (rows == 0) return 0;
        int cols = A[0].size();
        long long best = LLONG_MIN;
        for (int left = 0; left < cols; ++left) {
            std::vector<long long> tmp(rows, 0);
            for (int right = left; right < cols; ++right) {
                for (int r = 0; r < rows; ++r) tmp[r] += A[r][right];
                // Kadane on tmp
                long long cur = tmp[0], curBest = tmp[0];
                for (int i = 1; i < rows; ++i) {
                    cur = std::max(tmp[i], cur + tmp[i]);
                    curBest = std::max(curBest, cur);
                }
                best = std::max(best, curBest);
            }
        }
        return best;
    }
};
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package problems

func MaxSumSubmatrix(A [][]int) int {
        rows := len(A)
        if rows == 0 { return 0 }
        cols := len(A[0])
        best := -1 << 63
        for left := 0; left < cols; left++ {
                tmp := make([]int, rows)
                for right := left; right < cols; right++ {
                        for r := 0; r < rows; r++ { tmp[r] += A[r][right] }
                        // Kadane
                        cur := tmp[0]
                        curBest := tmp[0]
                        for i := 1; i < rows; i++ {
                                if cur + tmp[i] > tmp[i] { cur = cur + tmp[i] } else { cur = tmp[i] }
                                if cur > curBest { curBest = cur }
                        }
                        if curBest > best { best = curBest }
                }
        }
        return best
}
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class Solution {
    public long maxSumSubmatrix(int[][] A) {
        int rows = A.length;
        if (rows == 0) return 0;
        int cols = A[0].length;
        long best = Long.MIN_VALUE;
        for (int left = 0; left < cols; left++) {
            long[] tmp = new long[rows];
            for (int right = left; right < cols; right++) {
                for (int r = 0; r < rows; r++) tmp[r] += A[r][right];
                long cur = tmp[0], curBest = tmp[0];
                for (int i = 1; i < rows; i++) {
                    cur = Math.max(tmp[i], cur + tmp[i]);
                    curBest = Math.max(curBest, cur);
                }
                best = Math.max(best, curBest);
            }
        }
        return best;
    }
}
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from typing import List

class Solution:
        def max_sum_submatrix(self, A: List[List[int]]) -> int:
                rows = len(A)
                if rows == 0:
                        return 0
                cols = len(A[0])
                best = float('-inf')
                for left in range(cols):
                        tmp = [0] * rows
                        for right in range(left, cols):
                                for r in range(rows):
                                        tmp[r] += A[r][right]
                                cur = tmp[0]
                                cur_best = tmp[0]
                                for x in tmp[1:]:
                                        cur = max(x, cur + x)
                                        cur_best = max(cur_best, cur)
                                best = max(best, cur_best)
                return best

Complexity

  • ⏰ Time complexity: O(cols^2 * rows) — we iterate over all column pairs (O(cols^2)) and for each pair run Kadane over rows.
  • 🧺 Space complexity: O(rows) — temporary array tmp of length rows.