0829HardArrayTwo PointersBreadth-First SearchUnion-FindSortingHeap (Priority Queue)Matrix

Cells Cleared Under a Depth Budget

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Trains the technique from

LeetCode 2503Maximum Number of Points From Grid Queries

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 survey grid records the silt depth of each cell, given as grid with m rows and n columns.

A dredger works one budget at a time. For a budget b it starts at the cell in row 0, column 0 and repeatedly moves to a cell sharing an edge with a cell it has already cleared. A cell may be cleared only when its recorded depth is strictly less than b, and that includes the starting cell: if the starting cell is not shallow enough, nothing is cleared at all. Each cell is counted once however often it is visited.

queries[i] holds the budget for the i-th run, and the runs are independent of each other. Return an array cleared where cleared[i] is how many cells the i-th run can clear.

Examples

Example 1

Input: grid = [[1, 2, 3], [2, 5, 7], [3, 5, 1]], queries = [5, 6, 2]
Output: [5, 8, 1]

Under a budget of 5 the dredger clears the cells recording 1, 2, 3, 2 and 3, which is five cells. Under a budget of 2 only the opening cell recording 1 is shallow enough, so one cell is cleared.

Example 2

Input: grid = [[5, 2], [1, 1]], queries = [3]
Output: [0]

The opening cell records a depth of 5, which is not below the budget of 3, so nothing at all is cleared.

Example 3

Input: grid = [[1, 2, 9], [9, 3, 9], [5, 4, 9]], queries = [6]
Output: [5]

The dredger clears 1, then 2 to its right, then 3 below that, then 4 below that, then 5 to the left of the 4, which is five cells.

Constraints

2 <= grid.length <= 1000
2 <= grid[0].length <= 1000
Every row of grid has the same length
1 <= queries.length <= 10^4
1 <= grid[i][j] <= 10^6
1 <= queries[i] <= 10^6

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 cells_cleared(grid: list[list[int]], queries: list[int]) -> list[int]:

Java

public int[] cellsCleared(int[][] grid, int[] queries)

The intended approach

Time O(m * n log(m * n) + q log q) for a grid of m by n and q budgets.Space O(m * n) for the frontier heap and the touched marks.

A larger budget clears everything a smaller one does, so answering the budgets in increasing order lets a single flood keep growing rather than restarting. Hold the frontier in a heap keyed on depth. For each budget in turn, pop and open every frontier cell whose depth is below it, pushing each newly touched neighbour, and record the running opened count against that budget's original slot. Each cell enters the heap once across the whole run, so the answers come back in the order they were asked despite being computed in sorted order.

Pythonentry point: cells_cleared

def cells_cleared(grid, queries):
    import heapq
    rows = len(grid)
    cols = len(grid[0])
    # answer the budgets in increasing order and keep one growing flood, so each
    # cell is opened at most once across all of them
    order = sorted(range(len(queries)), key=lambda i: queries[i])
    out = [0] * len(queries)
    seen = [[False] * cols for _ in range(rows)]
    seen[0][0] = True
    edge = [(grid[0][0], 0, 0)]
    opened = 0
    for slot in order:
        budget = queries[slot]
        while edge and edge[0][0] < budget:
            depth, r, c = heapq.heappop(edge)
            opened += 1
            for dr, dc in ((1, 0), (-1, 0), (0, 1), (0, -1)):
                nr = r + dr
                nc = c + dc
                if 0 <= nr < rows and 0 <= nc < cols and not seen[nr][nc]:
                    seen[nr][nc] = True
                    heapq.heappush(edge, (grid[nr][nc], nr, nc))
        out[slot] = opened
    return out

The obvious version

Run each budget from scratch: flood out from the opening cell with an explicit stack, admitting a neighbour only when its depth is below the budget, and count what was reached. Repeats the whole flood once per budget.

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_cells_cleared(grid, queries):
    rows = len(grid)
    cols = len(grid[0])
    out = []
    for budget in queries:
        if grid[0][0] >= budget:
            out.append(0)
            continue
        seen = [[False] * cols for _ in range(rows)]
        seen[0][0] = True
        stack = [(0, 0)]
        count = 0
        while stack:
            r, c = stack.pop()
            count += 1
            for dr, dc in ((1, 0), (-1, 0), (0, 1), (0, -1)):
                nr = r + dr
                nc = c + dc
                if (0 <= nr < rows and 0 <= nc < cols and not seen[nr][nc]
                        and grid[nr][nc] < budget):
                    seen[nr][nc] = True
                    stack.append((nr, nc))
        out.append(count)
    return out

How this problem was checked

Reference solution executed against 12 test cases — 12 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

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