【LeetCode】947. Most Stones Removed with Same Row or Column (Medium) (JAVA)
Title address: https://leetcode.com/problems/minimum-increment-to-make-array-unique/
Title 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 <= 10^4
- No two stones are at the same coordinate point.
General idea
n stones are placed on some integer coordinate points in a two-dimensional plane. There can be at most one stone on each coordinate point.
If there are other stones in the same row or in the same row of a stone, then the stone can be removed.
Give you an array of stones of length n, where stones[i] = [xi, yi] represents the position of the i-th stone, and returns the maximum number of stones that can be removed.
Problem-solving method
- Adopt and find
- If x does not appear in the set, add one to the number; associate x and y in [x, y], and reduce the number by one;
- Note: Because the maximum number of x and y axes is uncertain, the method of map is used to store the father node of x; because xi, yi <= 10^4, 10001 can be added to xi and yi to avoid x and y repeat
class Solution {
public int removeStones(int[][] stones) {
UnionFind unionFind = new UnionFind();
for (int i = 0; i < stones.length; i++) {
unionFind.merge(stones[i][0], stones[i][1] + 10001);
}
return stones.length - unionFind.getCount();
}
public class UnionFind {
private Map<Integer, Integer> map = new HashMap<>();
private int count = 0;
public int getCount() {
return count;
}
public int find(int x) {
if (!map.containsKey(x)) {
map.put(x, x);
count++;
}
int fatherX = map.get(x);
if (fatherX != x) {
map.put(x, find(fatherX));
}
return map.get(x);
}
public void merge(int x, int y) {
int fatherX = find(x);
int fatherY = find(y);
if (fatherX == fatherY) return;
map.put(fatherX, fatherY);
count--;
}
}
}
Execution time: 9 ms, beating 69.65% of Java users
Memory consumption: 38.9 MB, beating 35.53% of Java users
