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Most Stones Removed with Same Row or Column⚓︎

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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] = [x_i, y_i] represents the location of the i-th 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:
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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:
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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 <= x_i, y_i <= 10^4
  • No two stones are at the same coordinate point.

Solution⚓︎

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class Solution {
private:
    unordered_map<int, int> rowFirst, columnFirst;
    vector<int> p;
    int sets;

    void build(int n) {
        for (int i = 0; i < n; i++) p.emplace_back(i);
        sets = n;
    }

    int find(int i) {
        if (p[i] != i) p[i] = find(p[i]);
        return p[i];
    }

    void unionSets(int x, int y) {
        int fx = find(x), fy = find(y);
        if (fx != fy) {
            p[fx] = fy;
            sets--;
        }
    }

public:
    int removeStones(vector<vector<int>>& stones) {
        int n = stones.size();
        build(n);

        for (int i = 0; i < n; i++) {
            int row = stones[i][0], col = stones[i][1];
            if (!rowFirst.count(row))
                rowFirst[row] = i;
            else
                unionSets(i, rowFirst[row]);

            if (!columnFirst.count(col))
                columnFirst[col] = i;
            else
                unionSets(i, columnFirst[col]);
        }

        return n - sets;
    }
};
  • Time complexity: \(O(n\times \alpha (n))\);
  • Space complexity: \(O(n)\).