Most Stones Removed with Same Row or Column

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 <= 104
• No two stones are at the same coordinate point.

Code

Union and Find

Time Complexity: $O(n)$, Space Complexity: $O(n)$

對 example 進行實作就會發現，要求的就是總共的 stones 數量減去 Number of Islands 。

另一個關鍵在於該如何對 (x, y) 座標進行 union and find，在這裡我們對 y 進行 encoding，y -> ~y，因為 y + ~y = -1，因此 ~y = -1 - y。如此處理後可以避免 x, y 具有重複的 range 造成我們在 union and find 時會無法區分到底是 row index 還是 column index。

class Solution {
    int island = 0;
    unordered_map<int, int> parent;
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() - island;
    }
 
    void uni(int x, int y) {
        int px = find(x), py = find(y);
        if(px != py) {
            parent[px] = py;
            island--;
        }
    }
 
    int find(int x) {
        if(!parent.count(x)) {
            parent[x] = x;
            island++;
        } 
        
        if(parent[x] != x) {
            parent[x] = find(parent[x]);
        }
        return parent[x];
 
    }
 
 
};

Source

• Most Stones Removed with Same Row or Column - LeetCode