- Difficulty: Medium
- Tags: LeetCode, Medium, Depth-First Search, Union Find, Graph, Hash Table, leetcode-947, Array, O(n)
Problem
On a 2D plane, we place n
stones at some integer coordinate points. Each coordinate point may have at most one stone.
A stone can be removed if it shares either the same row or the same column as another stone that has not been removed.
Given an array stones
of length n
where stones[i] = [xi, yi]
represents the location of the ith
stone, return the largest possible number of stones that can be removed.
Example 1:
Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]] Output: 5 Explanation: One way to remove 5 stones is as follows: 1. Remove stone [2,2] because it shares the same row as [2,1]. 2. Remove stone [2,1] because it shares the same column as [0,1]. 3. Remove stone [1,2] because it shares the same row as [1,0]. 4. Remove stone [1,0] because it shares the same column as [0,0]. 5. Remove stone [0,1] because it shares the same row as [0,0]. Stone [0,0] cannot be removed since it does not share a row/column with another stone still on the plane.
Example 2:
Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]] Output: 3 Explanation: One way to make 3 moves is as follows: 1. Remove stone [2,2] because it shares the same row as [2,0]. 2. Remove stone [2,0] because it shares the same column as [0,0]. 3. Remove stone [0,2] because it shares the same row as [0,0]. Stones [0,0] and [1,1] cannot be removed since they do not share a row/column with another stone still on the plane.
Example 3:
Input: stones = [[0,0]] Output: 0 Explanation: [0,0] is the only stone on the plane, so you cannot remove it.
Constraints:
1 <= stones.length <= 1000
0 <= xi, yi <= 104
- No two stones are at the same coordinate point.