Date: 2026-06-06
Time: 16:15
This file solves LeetCode 2500: Delete Greatest Value in Each Row. It implements the simulation where you repeatedly delete the maximum value from each row of a matrix, take the max of those deleted values, and sum those maxima across all rounds. The file owns both the solution and its unit tests.
maxValueAfterOperations(grid: List[List[int]]) -> intThe core function. It exploits an insight that avoids actually simulating deletions: if you sort each row, then column j across all rows contains exactly the values that would be deleted together in operation j (from smallest to largest). The answer is just the sum of column-wise maximums over the sorted grid.
Contract: Takes an m x n grid of positive integers, returns an integer. Mutates grid in place (sorts each row).
TestMaxValueAfterOperationsSeven test cases covering: the two LeetCode examples, single-row, single-column, uniform values, descending rows, and a 50x50 uniform grid for scale.
Sort-then-reduce: Rather than simulating O(n) deletion rounds with O(m) max-finding each, the solution sorts each row once and reduces to column-max summation. This is a common LeetCode idiom — transform the structure so the answer falls out of a simpler traversal.
In-place mutation: row.sort() modifies the input grid. This is fine for LeetCode but worth noting — callers lose the original ordering.
Generator inside sum: The return statement uses a nested generator expression: the outer iterates columns, the inner finds the max across rows for that column. No intermediate lists are allocated.
Imports: unittest (test harness), typing.List (type annotation). No external dependencies.
Imported by: The test_solution.py in the same directory imports this. The massive "Imported By" list in the prompt is misleading — those are unrelated test files across different problem directories that happen to share a similar import structure, not actual consumers of this function.
1. Sort every row of grid in ascending order — O(m * n log n)
2. For each column index j from 0 to n-1, compute max(grid[i][j] for i in range(m))
3. Sum those column maxima and return
Why this works: Sorting aligns the k-th smallest element of each row into column k. The problem asks you to delete the largest remaining element from each row simultaneously, which after sorting is column n-1 first, then n-2, etc. Since we sum all column maxima regardless of order, the traversal direction doesn't matter.
grid[i][j] <= grid[i][j+1] for all valid i, jgrid is non-empty and rectangular (all rows have equal length) — no validationNone. An empty grid or jagged rows would raise IndexError or ValueError from range(len(grid[0])) or max(). This is standard for LeetCode solutions where input constraints guarantee well-formed data.
delete-greatest-value-in-each-row/review.md — The code review may document alternative approaches or edge-case analysisarray-partition/solution.py:arrayPairSum — Uses the same sort-then-sum-by-position pattern for a related greedy problemsort-as-simulation-substitute — Across this repo, many solutions replace an iterative simulation with a single sort; identifying which problems admit this is a key skilllucky-numbers-in-a-matrix/solution.py — Another matrix problem that combines row and column extrema, worth comparing the reduction approachsort-then-column-max-equivalence — Sorting each row and summing column-wise maxima produces the same result as simulating repeated deletion of row maximain-place-grid-mutation — maxValueAfterOperations mutates the input grid by sorting each row; the original order is destroyedno-input-validation — The function assumes a non-empty rectangular grid and will raise on empty or jagged inputtime-complexity-n-log-n — The dominant cost is sorting m rows of length n, giving O(m * n log n) total time