LeetCode #1975 — MEDIUM

Maximum Matrix Sum

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an n x n integer matrix. You can do the following operation any number of times:

Choose any two adjacent elements of matrix and multiply each of them by -1.

Two elements are considered adjacent if and only if they share a border.

Your goal is to maximize the summation of the matrix's elements. Return the maximum sum of the matrix's elements using the operation mentioned above.

Example 1:

Input: matrix = [[1,-1],[-1,1]]
Output: 4
Explanation: We can follow the following steps to reach sum equals 4:
- Multiply the 2 elements in the first row by -1.
- Multiply the 2 elements in the first column by -1.

Example 2:

Input: matrix = [[1,2,3],[-1,-2,-3],[1,2,3]]
Output: 16
Explanation: We can follow the following step to reach sum equals 16:
- Multiply the 2 last elements in the second row by -1.

Constraints:

n == matrix.length == matrix[i].length
2 <= n <= 250
-10⁵ <= matrix[i][j] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an n x n integer matrix. You can do the following operation any number of times: Choose any two adjacent elements of matrix and multiply each of them by -1. Two elements are considered adjacent if and only if they share a border. Your goal is to maximize the summation of the matrix's elements. Return the maximum sum of the matrix's elements using the operation mentioned above.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[[1,-1],[-1,1]]

Example 2

[[1,2,3],[-1,-2,-3],[1,2,3]]

Step 02

Core Insight

What unlocks the optimal approach

Try to use the operation so that each row has only one negative number.
If you have only one negative element you cannot convert it to positive.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1975: Maximum Matrix Sum
class Solution {
    public long maxMatrixSum(int[][] matrix) {
        long s = 0;
        int mi = 1 << 30, cnt = 0;
        for (var row : matrix) {
            for (int x : row) {
                cnt += x < 0 ? 1 : 0;
                int y = Math.abs(x);
                mi = Math.min(mi, y);
                s += y;
            }
        }
        return cnt % 2 == 0 ? s : s - mi * 2;
    }
}

// Accepted solution for LeetCode #1975: Maximum Matrix Sum
func maxMatrixSum(matrix [][]int) int64 {
	var s int64
	mi, cnt := 1<<30, 0
	for _, row := range matrix {
		for _, x := range row {
			if x < 0 {
				cnt++
				x = -x
			}
			mi = min(mi, x)
			s += int64(x)
		}
	}
	if cnt%2 == 0 {
		return s
	}
	return s - int64(mi*2)
}

# Accepted solution for LeetCode #1975: Maximum Matrix Sum
class Solution:
    def maxMatrixSum(self, matrix: List[List[int]]) -> int:
        mi = inf
        s = cnt = 0
        for row in matrix:
            for x in row:
                cnt += x < 0
                y = abs(x)
                mi = min(mi, y)
                s += y
        return s if cnt % 2 == 0 else s - mi * 2

// Accepted solution for LeetCode #1975: Maximum Matrix Sum
impl Solution {
    pub fn max_matrix_sum(matrix: Vec<Vec<i32>>) -> i64 {
        let mut s = 0;
        let mut mi = i32::MAX;
        let mut cnt = 0;
        for row in matrix {
            for &x in row.iter() {
                cnt += if x < 0 { 1 } else { 0 };
                let y = x.abs();
                mi = mi.min(y);
                s += y as i64;
            }
        }
        if cnt % 2 == 0 {
            s
        } else {
            s - (mi as i64 * 2)
        }
    }
}

// Accepted solution for LeetCode #1975: Maximum Matrix Sum
function maxMatrixSum(matrix: number[][]): number {
    let [s, cnt, mi] = [0, 0, Infinity];
    for (const row of matrix) {
        for (const x of row) {
            if (x < 0) {
                ++cnt;
            }
            const y = Math.abs(x);
            s += y;
            mi = Math.min(mi, y);
        }
    }
    return cnt % 2 === 0 ? s : s - 2 * mi;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.