Question

947. Most Stones Removed with Same Row or Column

中文文档

Description

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.

Solutions

Solution 1

class Solution:
  def removeStones(self, stones: List[List[int]]) -> int:
    def find(x):
      if p[x] != x:
        p[x] = find(p[x])
      return p[x]

    n = 10010
    p = list(range(n << 1))
    for x, y in stones:
      p[find(x)] = find(y + n)

    s = {find(x) for x, _ in stones}
    return len(stones) - len(s)

class Solution {
  private int[] p;

  public int removeStones(int[][] stones) {
    int n = 10010;
    p = new int[n << 1];
    for (int i = 0; i < p.length; ++i) {
      p[i] = i;
    }
    for (int[] stone : stones) {
      p[find(stone[0])] = find(stone[1] + n);
    }
    Set<Integer> s = new HashSet<>();
    for (int[] stone : stones) {
      s.add(find(stone[0]));
    }
    return stones.length - s.size();
  }

  private int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }
}

class Solution {
public:
  vector<int> p;

  int removeStones(vector<vector<int>>& stones) {
    int n = 10010;
    p.resize(n << 1);
    for (int i = 0; i < p.size(); ++i) p[i] = i;
    for (auto& stone : stones) p[find(stone[0])] = find(stone[1] + n);
    unordered_set<int> s;
    for (auto& stone : stones) s.insert(find(stone[0]));
    return stones.size() - s.size();
  }

  int find(int x) {
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
  }
};

func removeStones(stones [][]int) int {
  n := 10010
  p := make([]int, n<<1)
  for i := range p {
    p[i] = i
  }
  var find func(x int) int
  find = func(x int) int {
    if p[x] != x {
      p[x] = find(p[x])
    }
    return p[x]
  }
  for _, stone := range stones {
    p[find(stone[0])] = find(stone[1] + n)
  }
  s := make(map[int]bool)
  for _, stone := range stones {
    s[find(stone[0])] = true
  }
  return len(stones) - len(s)
}