LeetCode #2679 — MEDIUM

Sum in a Matrix

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

The Problem

Problem Statement

You are given a 0-indexed 2D integer array nums. Initially, your score is 0. Perform the following operations until the matrix becomes empty:

From each row in the matrix, select the largest number and remove it. In the case of a tie, it does not matter which number is chosen.
Identify the highest number amongst all those removed in step 1. Add that number to your score.

Return the final score.

Example 1:

Input: nums = [[7,2,1],[6,4,2],[6,5,3],[3,2,1]]
Output: 15
Explanation: In the first operation, we remove 7, 6, 6, and 3. We then add 7 to our score. Next, we remove 2, 4, 5, and 2. We add 5 to our score. Lastly, we remove 1, 2, 3, and 1. We add 3 to our score. Thus, our final score is 7 + 5 + 3 = 15.

Example 2:

Input: nums = [[1]]
Output: 1
Explanation: We remove 1 and add it to the answer. We return 1.

Constraints:

1 <= nums.length <= 300
1 <= nums[i].length <= 500
0 <= nums[i][j] <= 10³

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed 2D integer array nums. Initially, your score is 0. Perform the following operations until the matrix becomes empty: From each row in the matrix, select the largest number and remove it. In the case of a tie, it does not matter which number is chosen. Identify the highest number amongst all those removed in step 1. Add that number to your score. Return the final score.

Baseline thinking

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

Pattern signal: Array

Example 1

[[7,2,1],[6,4,2],[6,5,3],[3,2,1]]

Example 2

[[1]]

Step 02

Core Insight

What unlocks the optimal approach

Sort the numbers in each row in decreasing order.
The answer is the summation of the max number in every column after sorting the rows.

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 #2679: Sum in a Matrix
class Solution {
    public int matrixSum(int[][] nums) {
        for (var row : nums) {
            Arrays.sort(row);
        }
        int ans = 0;
        for (int j = 0; j < nums[0].length; ++j) {
            int mx = 0;
            for (var row : nums) {
                mx = Math.max(mx, row[j]);
            }
            ans += mx;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2679: Sum in a Matrix
func matrixSum(nums [][]int) (ans int) {
	for _, row := range nums {
		sort.Ints(row)
	}
	for i := 0; i < len(nums[0]); i++ {
		mx := 0
		for _, row := range nums {
			mx = max(mx, row[i])
		}
		ans += mx
	}
	return
}

# Accepted solution for LeetCode #2679: Sum in a Matrix
class Solution:
    def matrixSum(self, nums: List[List[int]]) -> int:
        for row in nums:
            row.sort()
        return sum(map(max, zip(*nums)))

// Accepted solution for LeetCode #2679: Sum in a Matrix
impl Solution {
    pub fn matrix_sum(mut nums: Vec<Vec<i32>>) -> i32 {
        for row in &mut nums {
            row.sort();
        }
        (0..nums[0].len())
            .map(|col| nums.iter().map(|row| row[col]).max().unwrap())
            .sum()
    }
}

// Accepted solution for LeetCode #2679: Sum in a Matrix
function matrixSum(nums: number[][]): number {
    for (const row of nums) {
        row.sort((a, b) => a - b);
    }
    let ans = 0;
    for (let j = 0; j < nums[0].length; ++j) {
        let mx = 0;
        for (const row of nums) {
            mx = Math.max(mx, row[j]);
        }
        ans += mx;
    }
    return ans;
}

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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.