947. Most Stones Removed with Same Row or Column
Đề Bài
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 <= 104- No two stones are at the same coordinate point.
Thuật Toán & Kỹ Thuật
⏱️ Thời gian
O(n²)
💾 Không gian
O(n)
Lời Giải
C++
0947-most-stones-removed-with-same-row-or-column.cpp
class Solution {
public:
int root[1001];
int res = 0;
int findRoot(int v) {
if (v == root[v]) {
return v;
}
return root[v] = findRoot(root[v]);
}
void unionSet(int u, int v) {
u = findRoot(u);
v = findRoot(v);
if (u != v) {
root[v] = u;
res++;
}
}
int removeStones(vector<vector<int>>& stones) {
int n = stones.size();
for (int i = 0; i < n; ++i) {
root[i] = i;
}
for (int i = 0; i < n - 1; ++i) {
int x1 = stones[i][0];
int y1 = stones[i][1];
for (int j = i + 1; j < n; ++j) {
int x2 = stones[j][0];
int y2 = stones[j][1];
if (x1 == x2 || y1 == y2) {
unionSet(i, j);
}
}
}
return res;
}
};