Today I have done leetcode's August LeetCoding Challenge with cpp.

August LeetCoding Challenge 29

Description

Most Stones Removed with Same Row or Column

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.

Solution

class Solution {
  int sz = 0;
  unordered_map<int, int> parent;
  int find(int x) {
    if(parent.count(x) && parent[x] != x) {
      parent[x] = find(parent[x]);
    } else if (!parent.count(x)) {
      parent[x] = x;
      sz += 1;
    }
    return parent[x];
  }
  void merge(int x, int y) {
    x = find(x);
    y = find(y);
    if(x != y) {
      parent[x] = y;
      sz -= 1;
    }
  }
public:
  int removeStones(vector<vector<int>>& stones) {
    for(const auto &stone : stones) {
      merge(stone[0], ~stone[1]);
    }
    return stones.size() - sz;
  }
};