947 Most Stones Removed with Same Row or Column.py

Author: rajeevranjancom

947 Most Stones Removed with Same Row or Column.py

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+#!/usr/bin/python3
+"""
+On a 2D plane, we place stones at some integer coordinate points.  Each coordinate point may have at most one stone.
+
+Now, a move consists of removing a stone that shares a column or row with another stone on the grid.
+
+What is the largest possible number of moves we can make?
+
+
+
+Example 1:
+
+Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
+Output: 5
+Example 2:
+
+Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
+Output: 3
+Example 3:
+
+Input: stones = [[0,0]]
+Output: 0
+
+
+Note:
+
+1 <= stones.length <= 1000
+0 <= stones[i][j] < 10000
+"""
+from typing import List
+from collections import defaultdict
+
+
+class Solution:
+    def removeStones(self, stones: List[List[int]]) -> int:
+        """
+        convert to graph problem
+        each component in the graph can be removed to only one node
+        N - #component
+
+        construct graph O(N^2)
+        DFS - O(N)
+        """
+        G = defaultdict(list)
+        n = len(stones)
+        for i in range(n):
+            for j in range(i):
+                if stones[i][0] == stones[j][0] or stones[i][1] == stones[j][1]:
+                    G[i].append(j)
+                    G[j].append(i)
+
+        # dfs
+        comp_cnt = 0
+        visited = [False for _ in range(n)]
+        for i in range(n):
+            if not visited[i]:
+                comp_cnt += 1
+                self.dfs(G, i, visited)
+
+        return n - comp_cnt
+
+    def dfs(self, G, i, visited):
+        visited[i] = True
+        for nbr in G[i]:
+            if not visited[nbr]:
+                self.dfs(G, nbr, visited)
+
+
+if __name__ == "__main__":
+    assert Solution().removeStones([[0,0],[0,2],[1,1],[2,0],[2,2]]) == 3