Python Approach by using Euclidean distance and DFS
\n# Code\n\nclass Solution:\n    # DFS ...!\n    def dfs(self, i):\n        self.visit[i] = 1\n        \n        for node in self.d[i]:\n            if self.vis
\n# Code\n```\nclass Solution:\n    # DFS ...!\n    def dfs(self, i):\n        self.visit[i] = 1\n        \n        for node in self.d[i]:\n            if self.visit[node] == 1:\n                continue\n            self.dfs(node)\n\n    def maximumDetonation(self, bombs: List[List[int]]) -> int:\n        n = len(bombs)\n        d = {}\n        for i in range(n): d[i] = set()\n\n        for i in range(n):\n            for j in range(n):\n                if i == j: continue\n\n                # Euclidean Distance\n                temp =(bombs[i][0]-bombs[j][0])**2+(bombs[i][1]-bombs[j][1])**2      \n\n                if temp <= bombs[j][2]**2: d[j].add(i)\n                if temp <= bombs[i][2]**2: d[i].add(j)\n        \n        self.d = d\n        ans = 1\n        for i in range(n):\n            self.visit = [0]*n\n            self.dfs(i)\n            ans = max(ans , sum(self.visit))    # No.of 1\'s\n\n        return ans\n```
['Python3']