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 <= 10000 <= xi, yi <= 104class Solution {
public:
int removeStones(vector<vector<int>>& stones) {
for (int i = 0; i < stones.size(); ++i)
uni(stones[i][0], ~stones[i][1]);
return stones.size() - islands;
}
unordered_map<int, int> f;
int islands = 0;
int find(int x) {
if (!f.count(x)) f[x] = x, islands++;
if (x != f[x]) f[x] = find(f[x]);
return f[x];
}
void uni(int x, int y) {
x = find(x), y = find(y);
if (x != y) f[x] = y, islands--;
}
};class Solution {
private class UnionFind {
int setCount;
int[] root;
int[] rank;
UnionFind(int size) {
this.root = new int[size];
this.rank = new int[size];
this.setCount = size;
for (int i = 0; i < size; i++) {
root[i] = i;
rank[i] = 1;
}
}
int find(int x) {
if (x == root[x])
return x;
return root[x] = find(root[x]);
}
void union(int x, int y) {
int rootX = find(x);
int rootY = find(y);
if (rootX != rootY) {
if (rank[rootX] > rank[rootY])
root[rootY] = rootX;
else if (rank[rootX] < rank[rootY])
root[rootX] = rootY;
else {
root[rootX] = rootY;
rank[rootY]++;
}
this.setCount--;
}
}
boolean connected(int x, int y) {
return find(x) == find(y);
}
}
private int getId(int r, int c, int m, int n) {
return r * n + c;
}
private int getId(int[] coord, int m, int n) {
return coord[0] * n + coord[1];
}
public int removeStones(int[][] stones) {
int n = stones.length;
UnionFind uf = new UnionFind(n);
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
if (stones[i][0] == stones[j][0] || stones[i][1] == stones[j][1])
uf.union(i, j);
return n - uf.setCount;
}
}