0394HardArrayHash TableUnion-Find

Reclaimed Flats After Each Dump

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 survey crew is reclaiming a shallow tidal basin drawn as an m x n grid of cells. Every cell is open water on day zero.

You are handed positions, and positions[k] = [r, c] names the single cell that the barge packs with gravel on day k, which makes that cell dry. Should the crew pick a cell that is already dry, nothing about the basin changes that day.

Call two dry cells part of the same flat when they meet along a shared horizontal or vertical edge. Cells that only touch at a corner stay in separate flats.

Return answer, a list as long as positions, in which answer[k] records how many separate flats the basin holds once day k's load of gravel has been placed.

Examples

Example 1

Input: m = 2, n = 4, positions = [[0,1],[1,3],[0,3],[0,2]]
Output: [1,2,2,1]

Day 0 leaves the single dry cell (0,1). Day 1 dries (1,3), which touches nothing dry, so there are two flats. Day 2 dries (0,3), which shares an edge with (1,3), keeping the total at two. Day 3 dries (0,2), which shares edges with both (0,1) and (0,3), so every dry cell now sits in one flat.

Example 2

Input: m = 1, n = 5, positions = [[0,4],[0,2],[0,0],[0,3]]
Output: [1,2,3,2]

The first three days dry three cells that are pairwise non-adjacent, giving counts 1, 2 and 3. Day 3 dries (0,3), which shares edges with (0,2) and (0,4), leaving the flats {(0,0)} and {(0,2),(0,3),(0,4)}.

Example 3

Input: m = 3, n = 3, positions = [[1,1],[1,1],[2,1],[0,1]]
Output: [1,1,1,1]

Day 1 repeats cell (1,1), which is already dry, so the basin is unchanged and the count stays at 1. Days 2 and 3 each dry a cell sharing an edge with (1,1).

Example 4

Input: m = 2, n = 2, positions = [[0,0],[1,1],[0,1]]
Output: [1,2,1]

After day 1 the dry cells (0,0) and (1,1) touch only at a corner, so they are two flats. Day 2 dries (0,1), which shares an edge with each of them, so the count falls to one.

Constraints

1 <= m, n, positions.length <= 10^4
1 <= m * n <= 10^4
positions[k].length == 2
0 <= positions[k][0] < m
0 <= positions[k][1] < n

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 num_islands2(m: int, n: int, positions: list[list[int]]) -> list[int]:

Java

public List<Integer> numIslands2(int m, int n, int[][] positions)

The intended approach

Time O(m * n + k), k = len(positions), treating path-compressed union-find as near-constant per operationSpace O(m * n + k) for the parent and weight arrays plus the returned list

Keep a disjoint-set forest over the m * n cell indices, with -1 marking a cell that is still open water, and keep a running count of live flats. Drying a fresh cell raises the count by one, then each of its four edge neighbours that is already dry is unioned in, and the count drops by one only when that union actually joined two different roots. A repeat position is detected by its non-negative parent entry and simply re-reports the current count.

Pythonentry point: num_islands2

def num_islands2(m, n, positions):
    # Time O(m * n + k * a(m * n)) with k = len(positions); the near-constant
    # a() factor comes from path-compressed finds, so this is effectively
    # O(m * n + k). Space O(m * n + k): two flat arrays plus the returned list.
    parent = [-1] * (m * n)
    weight = [0] * (m * n)
    live = 0
    answer = []

    def find(x):
        root = x
        while parent[root] != root:
            root = parent[root]
        while parent[x] != root:
            parent[x], x = root, parent[x]
        return root

    for r, c in positions:
        idx = r * n + c
        if parent[idx] != -1:
            answer.append(live)
            continue
        parent[idx] = idx
        weight[idx] = 1
        live += 1
        for dr, dc in ((-1, 0), (1, 0), (0, -1), (0, 1)):
            nr, nc = r + dr, c + dc
            if 0 <= nr < m and 0 <= nc < n:
                nb = nr * n + nc
                if parent[nb] == -1:
                    continue
                ra = find(idx)
                rb = find(nb)
                if ra == rb:
                    continue
                if weight[ra] < weight[rb]:
                    ra, rb = rb, ra
                parent[rb] = ra
                weight[ra] += weight[rb]
                live -= 1
        answer.append(live)
    return answer

The obvious version

Hold a boolean grid, and after every dump recount connected groups from scratch with a flood fill over all cells.

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_num_islands2(m, n, positions):
    land = [[False] * n for _ in range(m)]
    answer = []
    for r, c in positions:
        land[r][c] = True
        seen = [[False] * n for _ in range(m)]
        total = 0
        for i in range(m):
            for j in range(n):
                if land[i][j] and not seen[i][j]:
                    total += 1
                    seen[i][j] = True
                    stack = [(i, j)]
                    while stack:
                        x, y = stack.pop()
                        for a, b in ((x - 1, y), (x + 1, y), (x, y - 1), (x, y + 1)):
                            if 0 <= a < m and 0 <= b < n and land[a][b] and not seen[a][b]:
                                seen[a][b] = True
                                stack.append((a, b))
        answer.append(total)
    return answer

How this problem was checked

Reference solution executed against 8 test cases — 8 passed
4 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