947. Most Stones Removed with Same Row or Column

Problem Link

Description

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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.

Solution

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Python3

class UnionFind:
    def __init__(self):
        self._parent = {}
        self._size = {}
 
    def union(self, a, b):
        a, b = self.find(a), self.find(b)
        if a == b:
            return
        if self._size[a] < self._size[b]:
            a, b = b, a
        self._parent[b] = a
        self._size[a] += self._size[b]
 
    def find(self, x):
        if x not in self._parent:
            self._parent[x] = x
            self._size[x] = 1
        while self._parent[x] != x:
            self._parent[x] = self._parent[self._parent[x]]
            x = self._parent[x]
        return x
 
class Solution:
    def removeStones(self, stones: List[List[int]]) -> int:
        N = len(stones)
        uf = UnionFind()
        R = defaultdict(list)
        C = defaultdict(list)
 
        for i, (r, c) in enumerate(stones):
            R[r].append(i)
            C[c].append(i)
        
        for V in R.values():
            for i in range(1, len(V)):
                uf.union(V[i], V[i - 1])
 
        for V in C.values():
            for i in range(1, len(V)):
                uf.union(V[i], V[i - 1])       
 
        parents = set()
        for i in range(N):
            p = uf.find(i)
            parents.add(p)
        
        return N - len(parents)