Welcome to Subscribe On Youtube

947. 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.

Solutions

  • class Solution {
        private int[] p;
    
        public int removeStones(int[][] stones) {
            int n = 10010;
            p = new int[n << 1];
            for (int i = 0; i < p.length; ++i) {
                p[i] = i;
            }
            for (int[] stone : stones) {
                p[find(stone[0])] = find(stone[1] + n);
            }
            Set<Integer> s = new HashSet<>();
            for (int[] stone : stones) {
                s.add(find(stone[0]));
            }
            return stones.length - s.size();
        }
    
        private int find(int x) {
            if (p[x] != x) {
                p[x] = find(p[x]);
            }
            return p[x];
        }
    }
    
  • class Solution {
    public:
        vector<int> p;
    
        int removeStones(vector<vector<int>>& stones) {
            int n = 10010;
            p.resize(n << 1);
            for (int i = 0; i < p.size(); ++i) p[i] = i;
            for (auto& stone : stones) p[find(stone[0])] = find(stone[1] + n);
            unordered_set<int> s;
            for (auto& stone : stones) s.insert(find(stone[0]));
            return stones.size() - s.size();
        }
    
        int find(int x) {
            if (p[x] != x) p[x] = find(p[x]);
            return p[x];
        }
    };
    
  • class Solution:
        def removeStones(self, stones: List[List[int]]) -> int:
            def find(x):
                if p[x] != x:
                    p[x] = find(p[x])
                return p[x]
    
            n = 10010
            p = list(range(n << 1))
            for x, y in stones:
                p[find(x)] = find(y + n)
    
            s = {find(x) for x, _ in stones}
            return len(stones) - len(s)
    
    
  • func removeStones(stones [][]int) int {
    	n := 10010
    	p := make([]int, n<<1)
    	for i := range p {
    		p[i] = i
    	}
    	var find func(x int) int
    	find = func(x int) int {
    		if p[x] != x {
    			p[x] = find(p[x])
    		}
    		return p[x]
    	}
    	for _, stone := range stones {
    		p[find(stone[0])] = find(stone[1] + n)
    	}
    	s := make(map[int]bool)
    	for _, stone := range stones {
    		s[find(stone[0])] = true
    	}
    	return len(stones) - len(s)
    }
    
  • class UnionFind {
        p: number[];
        size: number[];
        constructor(n: number) {
            this.p = Array.from({ length: n }, (_, i) => i);
            this.size = Array(n).fill(1);
        }
    
        find(x: number): number {
            if (this.p[x] !== x) {
                this.p[x] = this.find(this.p[x]);
            }
            return this.p[x];
        }
    
        union(a: number, b: number): boolean {
            const [pa, pb] = [this.find(a), this.find(b)];
            if (pa === pb) {
                return false;
            }
            if (this.size[pa] > this.size[pb]) {
                this.p[pb] = pa;
                this.size[pa] += this.size[pb];
            } else {
                this.p[pa] = pb;
                this.size[pb] += this.size[pa];
            }
            return true;
        }
    }
    
    function removeStones(stones: number[][]): number {
        const n = stones.length;
        const uf = new UnionFind(n);
        let ans = 0;
        for (let i = 0; i < n; ++i) {
            for (let j = 0; j < i; ++j) {
                if (stones[i][0] === stones[j][0] || stones[i][1] === stones[j][1]) {
                    ans += uf.union(i, j) ? 1 : 0;
                }
            }
        }
        return ans;
    }
    
    

All Problems

All Solutions