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Delete Greatest Value in Each Row — Easy Problem & Solution

Repeat until the grid is empty: delete the greatest value from each row (on a tie within a row, delete any one of them); add the largest of the deleted…

Problem statement

Repeat until the grid is empty:

  • delete the greatest value from each row (on a tie within a row, delete any one of them);
  • add the largest of the deleted values to your answer.

Return the final answer.

Example 1

Input: grid = [[1,2,4],[3,3,1]]
Output: 8
Explanation: Rounds delete `{4,3}`, `{2,3}` and `{1,1}`, adding 4 + 3 + 1.

Example 2

Input: grid = [[10]]
Output: 10

Example 3

Input: grid = [[1,2],[3,4]]
Output: 7
Explanation: Delete `{2,4}` for 4, then `{1,3}` for 3.

Constraints

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

How to solve Delete Greatest Value in Each Row

Sort every row. Round k then deletes exactly the k-th largest of each row — that is, one column of the sorted grid — so the answer is the sum of the column maxima.

Approach

  1. Sort each row ascending.
  2. For each column, take the maximum down the column.
  3. Sum those maxima.

Why it works

Sorting is what removes the simulation: rounds proceed in lockstep and a row always yields its next-largest remaining value, so which values meet in a round is fixed in advance by rank. Direction does not matter — summing column maxima of the ascending sort visits the same multiset of rounds as descending, just in the other order.

Complexity

  • Time — O(m · n log n)
  • Space — O(m · n)

Pitfalls

  • Each row loses one value per round, not just the global maximum.
  • Ties inside a row are irrelevant — the values are equal.
  • The answer accumulates one value per round, which is n in total.

Reference solution

Python

from typing import List

def deleteGreatestValue(grid: List[List[int]]) -> int:
    rows = [sorted(r) for r in grid]
    return sum(max(col) for col in zip(*rows))

JavaScript

var deleteGreatestValue = function(grid) {
    var rows = [];
    for (var i = 0; i < grid.length; i++) {
        var r = grid[i].slice();
        r.sort(function(a, b) { return a - b; });
        rows.push(r);
    }
    var m = rows.length, n = rows[0].length, total = 0;
    for (var col = 0; col < n; col++) {
        var best = 0;
        for (i = 0; i < m; i++) if (rows[i][col] > best) best = rows[i][col];
        total += best;
    }
    return total;
};

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