1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
|
// 947. Most Stones Removed with Same Row or Column
#include "leetcode.h"
/*
* 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] = [x_i,
* y_i]' represents the location of the i'th stone, return the largest possible
* number of stones that can be removed
*/
class Solution {
public:
int removeStones(vector<vector<int>> &stones) {
int n = stones.size(), c = 0;
unordered_map<int, vector<int>> rows, cols;
for (int i = 0; i < n; ++i) {
rows[stones[i][0]].push_back(i);
cols[stones[i][1]].push_back(i);
}
unordered_set<int> seen;
function<int(int)> dfs;
dfs = [&](int i) -> int {
if (seen.count(i))
return 0;
seen.insert(i);
int j = stones[i][0], k = stones[i][1];
for (auto r : rows[j])
dfs(r);
for (auto c : cols[k])
dfs(c);
return 1;
};
for (int i = 0; i < n; ++i)
c += dfs(i);
return n - c;
}
};
int main() {
Solution obj;
vvd(int) s1 = {{0, 0}, {0, 1}, {1, 0}, {1, 2}, {2, 1}, {2, 2}},
s2 = {{0, 1}, {0, 2}, {1, 1}, {2, 0}, {2, 2}};
printf("%d\n", obj.removeStones(s1)); // expect: 5
printf("%d\n", obj.removeStones(s2)); // expect: 3
}
|