🌟 Maximize Removal of Edges for Optimal Connectivity! 🛠️ Beats 97%✨
\n\n# Intuition\nThe problem involves maximizing the number of edges that can be removed from a graph while ensuring connectivity based on specified edge types.
![image.png](https://assets.leetcode.com/users/images/e05ca90b-47b3-4666-9f81-e7b59b89600c_1719718308.1541388.png)\n\n# Intuition\nThe problem involves maximizing the number of edges that can be removed from a graph while ensuring connectivity based on specified edge types. This requires understanding and manipulating the graph\'s connectivity using DFS (Depth-First Search) and component analysis.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Graph Representation: Represent the graph using adjacency lists for efficient traversal and\ncomponent identification.\n2. DFS for Component Identification: Use DFS to identify connected components of the graph\nand count the number of nodes in each component.\n3. Edge Removal Strategy: Separate edges based on their types (Type 1, Type 2, and Type 3) and\napply a strategy to maximize removals while maintaining connectivity:\n\u2022 For Type 3 edges, ensure all nodes in the same connected component.\n\u2022 For Type 1 and Type 2 edges, ensure connectivity between all nodes using these edges.\n# Detailed Steps\n1. Initialize Data Structures: Create adjacency lists for the graph and arrays to track visited nodes\nand connected components.\n2. DFS for Component Identification: Implement a DFS function to mark nodes with their\nrespective component numbers and count nodes in each component.\n3. Edge Categorization: Separate edges into Type 1, Type 2, and Type 3 based on their given\nspecifications.\n4. Ensure Connectivity: For Type 3 edges, ensure all nodes are in the same connected component.\nFor Type 1 and Type 2 edges, use DFS to ensure connectivity across the graph.\n5. Calculate Maximum Removals: Calculate the maximum number of edges that can be removed\nwhile maintaining connectivity and meeting the specified conditions.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n+m), where n is the number of nodes and m is the number of edges. This includes DFS operations and edge categorization.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n+m), for storing adjacency lists, visited arrays, and other data structures related to graph traversal.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n\n    def dfs(self,vis,li,com,i):\n        if(vis[i]):\n            return 0\n        vis[i]=com\n        ans=1\n        for j in li[i]:\n            ans+=self.dfs(vis,li,com,j)\n        return ans\n\n    def maxNumEdgesToRemove(self, n: int, e: List[List[int]]) -> int:\n        ans=0\n        li=[[] for i in range(n)]\n        for l in e:\n            if(l[0]==3):\n                li[l[1]-1].append(l[2]-1)\n                li[l[2]-1].append(l[1]-1)\n        vis=[0 for i in range(n)]\n        com=0\n        for i in range(n):\n            if(vis[i]):\n                continue\n            else:\n                com+=1\n                temp=self.dfs(vis,li,com,i)\n                ans+=(temp-1)\n        lia=[[] for i in range(com)]\n        lib=[[] for i in range(com)]\n        for l in e:\n            if(l[0]==1):\n                a=vis[l[1]-1]-1\n                b=vis[l[2]-1]-1\n                if(a==b):\n                    continue\n                lia[a].append(b)\n                lia[b].append(a)\n            elif(l[0]==2):\n                a=vis[l[1]-1]-1\n                b=vis[l[2]-1]-1\n                if(a==b):\n                    continue\n                lib[a].append(b)\n                lib[b].append(a)\n        visa=[0 for i in range(com)]\n        coma=0\n        comb=0\n        for i in range(com):\n            if(visa[i]):\n                continue\n            else:\n                coma+=1\n                if(coma>1):\n                    return -1\n                self.dfs(visa,lia,coma,i)\n        \n        visb=[0 for i in range(com)]\n        for i in range(com):\n            if(visb[i]):\n                continue\n            else:\n                comb+=1\n                if(comb>1):\n                    return -1\n                self.dfs(visb,lib,comb,i)\n        if(coma>1 or comb>1):\n            return -1\n        ans+=(com-1)*2\n        return len(e)-ans\n\n```\nIf you found this helpful, please upvote! \uD83D\uDC4D\n\n![main-qimg-a645ae706f7acacc929d906ca22011cd-lq.jpg](https://assets.leetcode.com/users/images/b79f0d83-2e4d-4691-8f7f-26da94c91ac1_1719718418.8351936.jpeg)\n
['Python3']