LeetCode #2500 — EASY

Delete Greatest Value in Each Row

Build confidence with an intuition-first walkthrough focused on array fundamentals.

Solve on LeetCode

The Problem

Problem Statement

You are given an m x n matrix grid consisting of positive integers.

Perform the following operation until grid becomes empty:

Delete the element with the greatest value from each row. If multiple such elements exist, delete any of them.
Add the maximum of deleted elements to the answer.

Note that the number of columns decreases by one after each operation.

Return the answer after performing the operations described above.

Example 1:

Input: grid = [[1,2,4],[3,3,1]]
Output: 8
Explanation: The diagram above shows the removed values in each step.
- In the first operation, we remove 4 from the first row and 3 from the second row (notice that, there are two cells with value 3 and we can remove any of them). We add 4 to the answer.
- In the second operation, we remove 2 from the first row and 3 from the second row. We add 3 to the answer.
- In the third operation, we remove 1 from the first row and 1 from the second row. We add 1 to the answer.
The final answer = 4 + 3 + 1 = 8.

Example 2:

Input: grid = [[10]]
Output: 10
Explanation: The diagram above shows the removed values in each step.
- In the first operation, we remove 10 from the first row. We add 10 to the answer.
The final answer = 10.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 50
1 <= grid[i][j] <= 100

Step 01

Brute Force Baseline

Problem summary: You are given an m x n matrix grid consisting of positive integers. Perform the following operation until grid becomes empty: Delete the element with the greatest value from each row. If multiple such elements exist, delete any of them. Add the maximum of deleted elements to the answer. Note that the number of columns decreases by one after each operation. Return the answer after performing the operations described above.

Baseline thinking

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

Pattern signal: Array

Example 1

[[1,2,4],[3,3,1]]

Example 2

[[10]]

Core Insight

What unlocks the optimal approach

Iterate from the first to the last row and if there exist some unmarked cells, take a maximum from them and mark that cell as visited.
Add a maximum of newly marked cells to answer and repeat that operation until the whole matrix becomes marked.

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 #2500: Delete Greatest Value in Each Row
class Solution {
    public int deleteGreatestValue(int[][] grid) {
        for (var row : grid) {
            Arrays.sort(row);
        }
        int ans = 0;
        for (int j = 0; j < grid[0].length; ++j) {
            int t = 0;
            for (int i = 0; i < grid.length; ++i) {
                t = Math.max(t, grid[i][j]);
            }
            ans += t;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2500: Delete Greatest Value in Each Row
func deleteGreatestValue(grid [][]int) (ans int) {
	for _, row := range grid {
		sort.Ints(row)
	}
	for j := range grid[0] {
		t := 0
		for i := range grid {
			if t < grid[i][j] {
				t = grid[i][j]
			}
		}
		ans += t
	}
	return
}

# Accepted solution for LeetCode #2500: Delete Greatest Value in Each Row
class Solution:
    def deleteGreatestValue(self, grid: List[List[int]]) -> int:
        for row in grid:
            row.sort()
        return sum(max(col) for col in zip(*grid))

// Accepted solution for LeetCode #2500: Delete Greatest Value in Each Row
impl Solution {
    pub fn delete_greatest_value(grid: Vec<Vec<i32>>) -> i32 {
        let mut grid = grid;
        for i in 0..grid.len() {
            grid[i].sort();
        }

        let mut ans = 0;
        for j in 0..grid[0].len() {
            let mut mx = 0;

            for i in 0..grid.len() {
                if grid[i][j] > mx {
                    mx = grid[i][j];
                }
            }

            ans += mx;
        }

        ans
    }
}

// Accepted solution for LeetCode #2500: Delete Greatest Value in Each Row
function deleteGreatestValue(grid: number[][]): number {
    for (const row of grid) {
        row.sort((a, b) => a - b);
    }

    let ans = 0;
    for (let j = 0; j < grid[0].length; ++j) {
        let t = 0;
        for (let i = 0; i < grid.length; ++i) {
            t = Math.max(t, grid[i][j]);
        }
        ans += t;
    }

    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(m × n × log n)

Space

O(log n)

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.