0380MediumHash TableDepth-First SearchUnion-FindGraph TheoryBipartite Graph

Pegboard Pin Removal

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Trains the technique from

LeetCode 947Most Stones Removed with Same Row or Column

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 large pegboard has pins pushed into some of its holes. pins[i] = [row_i, col_i] gives the hole holding pin i, and no two pins share a hole.

Pins come out one at a time. A pin may be pulled only if, at that moment, another pin still in the board sits in the same row or in the same column as it.

Return the largest number of pins that can be pulled out.

Examples

Example 1

Input: pins = [[0, 0], [0, 2], [2, 2], [2, 0], [1, 1]]
Output: 3

Pull [0, 2] (its row mate [0, 0] is still in), then [2, 2] (its column mate [2, 0] is still in), then [2, 0] (its column mate [0, 0] is still in). That leaves [0, 0] and [1, 1], which share neither a row nor a column, so three pins come out.

Example 2

Input: pins = [[0, 1], [1, 0]]
Output: 0

The two pins sit in different rows and different columns, so neither one is ever eligible to be pulled.

Example 3

Input: pins = [[0, 0], [0, 1], [1, 0], [1, 1]]
Output: 3

Pull [1, 1], then [1, 0], then [0, 1]; each had a row or column mate on the board at the time, and [0, 0] stays behind.

Example 4

Input: pins = [[3, 3]]
Output: 0

A lone pin has no row or column mate, so nothing can be pulled.

Constraints

1 <= pins.length <= 1000
0 <= row_i, col_i <= 10^4
No two pins occupy the same hole.

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 remove_stones(pins: list[list[int]]) -> int:

Java

public int removeStones(int[][] pins)

The intended approach

Time O(n * inverse-Ackermann(n))Space O(n)

Treat every row and every column that holds a pin as a node and let each pin join its row node to its column node in a disjoint-set structure. Pins whose rows and columns end up in the same set can be pulled down to a single survivor, so the answer is the pin count minus the number of distinct sets the pins land in. Rows and columns are keyed by tagged pairs, so row 4 and column 4 are never confused.

Pythonentry point: remove_stones

def remove_stones(pins):
    # Time O(n * inverse-Ackermann(n)) -- one union per pin, then one find per pin.
    # Space O(n) for the parent map over the rows and columns actually used.
    parent = {}

    def find(x):
        if x not in parent:
            parent[x] = x
        root = x
        while parent[root] != root:
            root = parent[root]
        while parent[x] != root:
            parent[x], x = root, parent[x]
        return root

    for row, col in pins:
        a = find(("row", row))
        b = find(("col", col))
        if a != b:
            parent[a] = b
    groups = set()
    for row, col in pins:
        groups.add(find(("row", row)))
    return len(pins) - len(groups)

The obvious version

Compare every pair of pins to build an explicit adjacency list, then count connected groups with a stack-driven walk and subtract that count from the number of pins. Quadratic and free of any disjoint-set bookkeeping.

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_remove_stones(pins):
    n = len(pins)
    adj = [[] for _ in range(n)]
    for i in range(n):
        for j in range(i + 1, n):
            if pins[i][0] == pins[j][0] or pins[i][1] == pins[j][1]:
                adj[i].append(j)
                adj[j].append(i)
    seen = [False] * n
    groups = 0
    for start in range(n):
        if seen[start]:
            continue
        groups += 1
        seen[start] = True
        stack = [start]
        while stack:
            node = stack.pop()
            for nxt in adj[node]:
                if not seen[nxt]:
                    seen[nxt] = True
                    stack.append(nxt)
    return n - groups

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

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