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1579. Remove Max Number of Edges to Keep Graph Fully Traversable

Description

Alice and Bob have an undirected graph of n nodes and three types of edges:

  • Type 1: Can be traversed by Alice only.
  • Type 2: Can be traversed by Bob only.
  • Type 3: Can be traversed by both Alice and Bob.

Given an array edges where edges[i] = [typei, ui, vi] represents a bidirectional edge of type typei between nodes ui and vi, find the maximum number of edges you can remove so that after removing the edges, the graph can still be fully traversed by both Alice and Bob. The graph is fully traversed by Alice and Bob if starting from any node, they can reach all other nodes.

Return the maximum number of edges you can remove, or return -1 if Alice and Bob cannot fully traverse the graph.

 

Example 1:

Input: n = 4, edges = [[3,1,2],[3,2,3],[1,1,3],[1,2,4],[1,1,2],[2,3,4]]
Output: 2
Explanation: If we remove the 2 edges [1,1,2] and [1,1,3]. The graph will still be fully traversable by Alice and Bob. Removing any additional edge will not make it so. So the maximum number of edges we can remove is 2.

Example 2:

Input: n = 4, edges = [[3,1,2],[3,2,3],[1,1,4],[2,1,4]]
Output: 0
Explanation: Notice that removing any edge will not make the graph fully traversable by Alice and Bob.

Example 3:

Input: n = 4, edges = [[3,2,3],[1,1,2],[2,3,4]]
Output: -1
Explanation: In the current graph, Alice cannot reach node 4 from the other nodes. Likewise, Bob cannot reach 1. Therefore it's impossible to make the graph fully traversable.

 

 

Constraints:

  • 1 <= n <= 105
  • 1 <= edges.length <= min(105, 3 * n * (n - 1) / 2)
  • edges[i].length == 3
  • 1 <= typei <= 3
  • 1 <= ui < vi <= n
  • All tuples (typei, ui, vi) are distinct.

Solutions

Solution 1

Thinking

Delete as many edges as possible while both Alice and Bob can still traverse the graph. \(n\) and the edge count reach \(10^5\). Type-\(3\) edges serve both and should be kept first; exclusive edges only patch each person's graph.

Two disjoint-set forests track the two reachability relations. Add type \(3\) first and discard a shared edge that would close a cycle. Then add types \(1\) and \(2\) the same way. A solution exists only when both forests have a single component.

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class UnionFind:
    def __init__(self, n):
        self.p = list(range(n))
        self.size = [1] * n
        self.cnt = n

    def find(self, x):
        if self.p[x] != x:
            self.p[x] = self.find(self.p[x])
        return self.p[x]

    def union(self, a, b):
        pa, pb = self.find(a - 1), self.find(b - 1)
        if pa == pb:
            return False
        if self.size[pa] > self.size[pb]:
            self.p[pb] = pa
            self.size[pa] += self.size[pb]
        else:
            self.p[pa] = pb
            self.size[pb] += self.size[pa]
        self.cnt -= 1
        return True


class Solution:
    def maxNumEdgesToRemove(self, n: int, edges: List[List[int]]) -> int:
        ufa = UnionFind(n)
        ufb = UnionFind(n)
        ans = 0
        for t, u, v in edges:
            if t == 3:
                if ufa.union(u, v):
                    ufb.union(u, v)
                else:
                    ans += 1
        for t, u, v in edges:
            if t == 1:
                ans += not ufa.union(u, v)
            if t == 2:
                ans += not ufb.union(u, v)
        return ans if ufa.cnt == 1 and ufb.cnt == 1 else -1

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class UnionFind {
    private int[] p;
    private int[] size;
    public int cnt;

    public UnionFind(int n) {
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            size[i] = 1;
        }
        cnt = n;
    }

    public int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

    public boolean union(int a, int b) {
        int pa = find(a - 1), pb = find(b - 1);
        if (pa == pb) {
            return false;
        }
        if (size[pa] > size[pb]) {
            p[pb] = pa;
            size[pa] += size[pb];
        } else {
            p[pa] = pb;
            size[pb] += size[pa];
        }
        --cnt;
        return true;
    }
}

class Solution {
    public int maxNumEdgesToRemove(int n, int[][] edges) {
        UnionFind ufa = new UnionFind(n);
        UnionFind ufb = new UnionFind(n);
        int ans = 0;
        for (var e : edges) {
            int t = e[0], u = e[1], v = e[2];
            if (t == 3) {
                if (ufa.union(u, v)) {
                    ufb.union(u, v);
                } else {
                    ++ans;
                }
            }
        }
        for (var e : edges) {
            int t = e[0], u = e[1], v = e[2];
            if (t == 1 && !ufa.union(u, v)) {
                ++ans;
            }
            if (t == 2 && !ufb.union(u, v)) {
                ++ans;
            }
        }
        return ufa.cnt == 1 && ufb.cnt == 1 ? ans : -1;
    }
}

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class UnionFind {
public:
    int cnt;

    UnionFind(int n) {
        p = vector<int>(n);
        size = vector<int>(n, 1);
        iota(p.begin(), p.end(), 0);
        cnt = n;
    }

    bool unite(int a, int b) {
        int pa = find(a - 1), pb = find(b - 1);
        if (pa == pb) {
            return false;
        }
        if (size[pa] > size[pb]) {
            p[pb] = pa;
            size[pa] += size[pb];
        } else {
            p[pa] = pb;
            size[pb] += size[pa];
        }
        --cnt;
        return true;
    }

    int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

private:
    vector<int> p, size;
};

class Solution {
public:
    int maxNumEdgesToRemove(int n, vector<vector<int>>& edges) {
        UnionFind ufa(n);
        UnionFind ufb(n);
        int ans = 0;
        for (auto& e : edges) {
            int t = e[0], u = e[1], v = e[2];
            if (t == 3) {
                if (ufa.unite(u, v)) {
                    ufb.unite(u, v);
                } else {
                    ++ans;
                }
            }
        }
        for (auto& e : edges) {
            int t = e[0], u = e[1], v = e[2];
            ans += t == 1 && !ufa.unite(u, v);
            ans += t == 2 && !ufb.unite(u, v);
        }
        return ufa.cnt == 1 && ufb.cnt == 1 ? ans : -1;
    }
};

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type unionFind struct {
    p, size []int
    cnt     int
}

func newUnionFind(n int) *unionFind {
    p := make([]int, n)
    size := make([]int, n)
    for i := range p {
        p[i] = i
        size[i] = 1
    }
    return &unionFind{p, size, n}
}

func (uf *unionFind) find(x int) int {
    if uf.p[x] != x {
        uf.p[x] = uf.find(uf.p[x])
    }
    return uf.p[x]
}

func (uf *unionFind) union(a, b int) bool {
    pa, pb := uf.find(a-1), uf.find(b-1)
    if pa == pb {
        return false
    }
    if uf.size[pa] > uf.size[pb] {
        uf.p[pb] = pa
        uf.size[pa] += uf.size[pb]
    } else {
        uf.p[pa] = pb
        uf.size[pb] += uf.size[pa]
    }
    uf.cnt--
    return true
}

func maxNumEdgesToRemove(n int, edges [][]int) (ans int) {
    ufa := newUnionFind(n)
    ufb := newUnionFind(n)
    for _, e := range edges {
        t, u, v := e[0], e[1], e[2]
        if t == 3 {
            if ufa.union(u, v) {
                ufb.union(u, v)
            } else {
                ans++
            }
        }
    }
    for _, e := range edges {
        t, u, v := e[0], e[1], e[2]
        if t == 1 && !ufa.union(u, v) {
            ans++
        }
        if t == 2 && !ufb.union(u, v) {
            ans++
        }
    }
    if ufa.cnt == 1 && ufb.cnt == 1 {
        return
    }
    return -1
}

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