0755EasyArrayMath

Trays Watered The Most Times

Tracked in this browser only

Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A greenhouse holds a rectangular block of seedling trays with rows rows and cols columns. Rows are numbered 1 to rows from the back wall forward, columns 1 to cols from the left wall rightward.

The irrigation arm runs a list of watering passes. Pass [a, b] sweeps the corner block anchored at row 1 and column 1: it waters every tray whose row number is at most a and whose column number is at most b, adding one to that tray's watering count. Every tray starts at a count of 0.

After all passes have run, return how many trays share the highest watering count in the greenhouse. If passes is empty, every tray is still tied on a count of 0.

Examples

Example 1

Input: rows = 4, cols = 5, passes = [[3, 2], [2, 4]]
Output: 4

The first pass waters rows 1 to 3 and columns 1 to 2, the second waters rows 1 to 2 and columns 1 to 4. Trays (1,1), (1,2), (2,1) and (2,2) end on a count of 2 and no tray reaches 3, so four trays share the highest count.

Example 2

Input: rows = 6, cols = 7, passes = [[2, 7], [6, 3]]
Output: 6

Trays in rows 1 to 2 and columns 1 to 3 are touched by both passes, giving them a count of 2, while every other watered tray sits at 1. That block holds six trays.

Example 3

Input: rows = 3, cols = 4, passes = []
Output: 12

No pass runs, so all twelve trays still hold a count of 0 and every one of them ties for the highest count.

Constraints

1 <= rows, cols <= 4 * 10^4
0 <= passes.length <= 10^4
passes[i].length == 2
1 <= passes[i][j] <= 4 * 10^4
Pass i is given as [a_i, b_i] with a_i at most rows and b_i at most cols.
Under these bounds the returned count is at most 1.6 * 10^9.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def max_watered_trays(rows: int, cols: int, passes: list[list[int]]) -> int:

Java

public int maxWateredTrays(int rows, int cols, int[][] passes)

The intended approach

Time O(p) for p passes: one walk over the list.Space O(1).

Every pass covers a block anchored at the same corner, so the trays covered by all of them are exactly those inside the tightest row extent and the tightest column extent seen. Track the smallest `a` and the smallest `b` over the list and multiply them. With no passes at all the whole block is tied on zero, so the answer is `rows * cols`.

Pythonentry point: max_watered_trays

def max_watered_trays(rows, cols, passes):
    # Time O(p) for p passes, space O(1).
    if not passes:
        return rows * cols
    top = rows
    left = cols
    for a, b in passes:
        if a < top:
            top = a
        if b < left:
            left = b
    return top * left

The obvious version

Materialise the whole tray grid, add one to every tray each pass touches, then scan the grid twice: once for the highest count and once to tally how many trays hold it. Costs rows * cols * passes work, so it is only usable on small greenhouses.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_max_watered(rows, cols, passes):
    grid = [[0] * cols for _ in range(rows)]
    for a, b in passes:
        for r in range(a):
            row = grid[r]
            for c in range(b):
                row[c] += 1
    best = 0
    for r in range(rows):
        for c in range(cols):
            if grid[r][c] > best:
                best = grid[r][c]
    total = 0
    for r in range(rows):
        for c in range(cols):
            if grid[r][c] == best:
                total += 1
    return total

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in