Maximizing Removable Edges While Ensuring Full Traversability in Undirected Graphs for Alice and Bob
Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires us to find the maximum number of edges that can be removed while e
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires us to find the maximum number of edges that can be removed while ensuring that the graph remains fully traversable by both Alice and Bob. The key insight is to use Union-Find (or Disjoint Set Union, DSU) to manage the connectivity of the graph for both Alice and Bob independently and together.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Union-Find Initialization:\n\n    We create two Union-Find data structures, one for Alice and one for Bob.\nWe also create a counter to track the number of removed edges.\n2. Processing Type 3 Edges:\n\n   These edges can be used by both Alice and Bob, so we attempt to union them in both Union-Find structures.\nIf a union operation fails (because the nodes are already connected), we increment the removed edges counter.\n3. Processing Type 1 and Type 2 Edges:\n\n   Type 1 edges are only for Alice, so we try to union them in Alice\'s Union-Find.\nType 2 edges are only for Bob, so we try to union them in Bob\'s Union-Find.\nAgain, if a union operation fails, we increment the removed edges counter.\n4. Validation:\n\n    After processing all edges, we check if both Union-Find structures have only one connected component (indicating full traversability).\nIf not, it means that either Alice or Bob (or both) cannot fully traverse the graph, and we return -1.\n# Complexity\n- Time complexity:The overall complexity is O(E\u22C5\u03B1(N)), where\nE is the number of edges and \u03B1 is the inverse Ackermann function, which is very slow-growing.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: The space complexity is O(N) due to the storage required for the Union-Find structures.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass UnionFind {\n    private int[] p;\n    private int[] size;\n    public int cnt;\n\n    public UnionFind(int n) {\n        p = new int[n];\n        size = new int[n];\n        for (int i = 0; i < n; ++i) {\n            p[i] = i;\n            size[i] = 1;\n        }\n        cnt = n;\n    }\n\n    public int find(int x) {\n        if (p[x] != x) {\n            p[x] = find(p[x]);\n        }\n        return p[x];\n    }\n\n    public boolean union(int a, int b) {\n        int pa = find(a - 1), pb = find(b - 1);\n        if (pa == pb) {\n            return false;\n        }\n        if (size[pa] > size[pb]) {\n            p[pb] = pa;\n            size[pa] += size[pb];\n        } else {\n            p[pa] = pb;\n            size[pb] += size[pa];\n        }\n        --cnt;\n        return true;\n    }\n}\n\nclass Solution {\n    public int maxNumEdgesToRemove(int n, int[][] edges) {\n        UnionFind ufa = new UnionFind(n);\n        UnionFind ufb = new UnionFind(n);\n        int ans = 0;\n        for (var e : edges) {\n            int t = e[0], u = e[1], v = e[2];\n            if (t == 3) {\n                if (ufa.union(u, v)) {\n                    ufb.union(u, v);\n                } else {\n                    ++ans;\n                }\n            }\n        }\n        for (var e : edges) {\n            int t = e[0], u = e[1], v = e[2];\n            if (t == 1 && !ufa.union(u, v)) {\n                ++ans;\n            }\n            if (t == 2 && !ufb.union(u, v)) {\n                ++ans;\n            }\n        }\n        return ufa.cnt == 1 && ufb.cnt == 1 ? ans : -1;\n    }\n}\n```
['Union Find', 'Graph', 'C++', 'Java']