What is the brute force approach and complexity for Most Stones Removed with Same Row or Column?

Link stones that share a row or column, count the groups with DFS, answer = stones − groups with time complexity O(N²) and space complexity O(N²).

What is the optimal approach and complexity for Most Stones Removed with Same Row or Column?

Each stone unites its row with its column; answer = stones − distinct roots with optimal time complexity O(N · α(N)) and space complexity O(N).

How do you identify the pattern for Most Stones Removed with Same Row or Column?

Look for key signals in the prompt: "same row or same column" indicates Two stones are linked through a shared line → make the rows and columns the DSU nodes.; "stone can be removed while another shares a line" indicates Each group must keep one stone, so the count is stones minus groups.; "largest number that can be removed" indicates Only the number of groups matters, not the order of removals..

Most Stones Removed with Same Row or Column

Medium·LeetCode #947 Pro OnlyMedium

LeetCode

Problem Statement

Stones sit on integer points of a 2D plane, with at most one stone per point. Each stone is given as [x, y]: its row is x and its column is y.

A stone can be removed when another stone that is still on the plane shares its row or its column. You may remove stones one at a time, following that rule.

Return the largest number of stones you can remove.

Input

stones, a list of [x, y] points, one per stone.

Output

One integer: the most stones that can be removed.

Example 1

Input: stones = [[0,0], [0,1], [1,0], [1,2], [2,1], [2,2]]

Output: 5

All six stones are linked through shared rows and columns, so they form one group. Five can be removed, and one must stay.

Example 2

Input: stones = [[0,0], [0,2], [1,1], [2,0], [2,2]]

Output: 3

The stones at (0,0), (0,2), (2,0) and (2,2) form one group, so three of them can go. The stone at (1,1) shares no row or column with another stone, so it stays.

Example 3

Input: stones = [[0,0]]

Output: 0

A single stone cannot be removed, because no other stone shares its row or column.

Constraints

• 1 ≤ stones.length ≤ 1000
• 0 ≤ x, y ≤ 10^4
• No two stones are at the same point.

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“same row or same column”

→ Two stones are linked through a shared line → make the rows and columns the DSU nodes.

“stone can be removed while another shares a line”

→ Each group must keep one stone, so the count is stones minus groups.

“largest number that can be removed”

→ Only the number of groups matters, not the order of removals.

Approach Complexity Ladder

Brute Force — compare every pair, then DFSO(N²)O(N²)

Link stones that share a row or column, count the groups with DFS, answer = stones − groups

Optimal — DSU over rows and columnsO(N · α(N))O(N)

Each stone unites its row with its column; answer = stones − distinct roots

Brute Force Intuition

Build a graph on the stones: two stones are neighbours when they share a row or a column. Count the connected groups with a DFS. Inside each group of k stones, k − 1 stones can be removed, so the answer is stones minus groups.

Recognition rule for this sub-pattern:

Start with n groups; each successful union reduces the count by one.

Remember this · Union-Find (DSU)

Trigger

Edges arrive one by one, "are these connected?", merge equivalent items, count components, extra edge creating a cycle.

Think

Am I repeatedly merging groups and asking whether two things are in the same group?

Key question

What are the elements, and what does it mean to union two of them?

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