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685. Redundant Connection II
Description
In this problem, a rooted tree is a directed graph such that, there is exactly one node (the root) for which all other nodes are descendants of this node, plus every node has exactly one parent, except for the root node which has no parents.
The given input is a directed graph that started as a rooted tree with n nodes (with distinct values from 1 to n), with one additional directed edge added. The added edge has two different vertices chosen from 1 to n, and was not an edge that already existed.
The resulting graph is given as a 2D-array of edges. Each element of edges is a pair [ui, vi] that represents a directed edge connecting nodes ui and vi, where ui is a parent of child vi.
Return an edge that can be removed so that the resulting graph is a rooted tree of n nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array.
Example 1:
Input: edges = [[1,2],[1,3],[2,3]] Output: [2,3]
Example 2:
Input: edges = [[1,2],[2,3],[3,4],[4,1],[1,5]] Output: [4,1]
Constraints:
n == edges.length3 <= n <= 1000edges[i].length == 21 <= ui, vi <= nui != vi
Solutions
Solution 1
Union find.
Solution 2: Union-Find (Template Approach)
Here is a template approach using Union-Find for your reference.
The time complexity is $O(n \alpha(n))$, and the space complexity is $O(n)$. Here, $n$ is the number of edges, and $\alpha(n)$ is the inverse Ackermann function, which can be considered a very small constant.
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class Solution { public int[] findRedundantDirectedConnection(int[][] edges) { int n = edges.length; int[] p = new int[n + 1]; for (int i = 0; i <= n; ++i) { p[i] = i; } UnionFind uf = new UnionFind(n + 1); int conflict = -1, cycle = -1; for (int i = 0; i < n; ++i) { int u = edges[i][0], v = edges[i][1]; if (p[v] != v) { conflict = i; } else { p[v] = u; if (!uf.union(u, v)) { cycle = i; } } } if (conflict == -1) { return edges[cycle]; } int v = edges[conflict][1]; if (cycle != -1) { return new int[] {p[v], v}; } return edges[conflict]; } } class UnionFind { public int[] p; public int n; public UnionFind(int n) { p = new int[n]; for (int i = 0; i < n; ++i) { p[i] = i; } this.n = n; } public boolean union(int a, int b) { int pa = find(a); int pb = find(b); if (pa == pb) { return false; } p[pa] = pb; --n; return true; } public int find(int x) { if (p[x] != x) { p[x] = find(p[x]); } return p[x]; } } // Solution 2 class UnionFind { private final int[] p; private final int[] size; 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; } } 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), pb = find(b); 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]; } return true; } } class Solution { public int[] findRedundantDirectedConnection(int[][] edges) { int n = edges.length; int[] ind = new int[n]; for (var e : edges) { ++ind[e[1] - 1]; } List<Integer> dup = new ArrayList<>(); for (int i = 0; i < n; ++i) { if (ind[edges[i][1] - 1] == 2) { dup.add(i); } } UnionFind uf = new UnionFind(n); if (!dup.isEmpty()) { for (int i = 0; i < n; ++i) { if (i == dup.get(1)) { continue; } if (!uf.union(edges[i][0] - 1, edges[i][1] - 1)) { return edges[dup.get(0)]; } } return edges[dup.get(1)]; } for (int i = 0;; ++i) { if (!uf.union(edges[i][0] - 1, edges[i][1] - 1)) { return edges[i]; } } } } -
class UnionFind { public: vector<int> p; int n; UnionFind(int _n) : n(_n) , p(_n) { iota(p.begin(), p.end(), 0); } bool unite(int a, int b) { int pa = find(a), pb = find(b); if (pa == pb) return false; p[pa] = pb; --n; return true; } int find(int x) { if (p[x] != x) p[x] = find(p[x]); return p[x]; } }; class Solution { public: vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) { int n = edges.size(); vector<int> p(n + 1); for (int i = 0; i <= n; ++i) p[i] = i; UnionFind uf(n + 1); int conflict = -1, cycle = -1; for (int i = 0; i < n; ++i) { int u = edges[i][0], v = edges[i][1]; if (p[v] != v) conflict = i; else { p[v] = u; if (!uf.unite(u, v)) cycle = i; } } if (conflict == -1) return edges[cycle]; int v = edges[conflict][1]; if (cycle != -1) return {p[v], v}; return edges[conflict]; } }; // Solution 2 class UnionFind { public: UnionFind(int n) { p = vector<int>(n); size = vector<int>(n, 1); iota(p.begin(), p.end(), 0); } bool unite(int a, int b) { int pa = find(a), pb = find(b); 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]; } 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: vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) { int n = edges.size(); vector<int> ind(n); for (const auto& e : edges) { ++ind[e[1] - 1]; } vector<int> dup; for (int i = 0; i < n; ++i) { if (ind[edges[i][1] - 1] == 2) { dup.push_back(i); } } UnionFind uf(n); if (!dup.empty()) { for (int i = 0; i < n; ++i) { if (i == dup[1]) { continue; } if (!uf.unite(edges[i][0] - 1, edges[i][1] - 1)) { return edges[dup[0]]; } } return edges[dup[1]]; } for (int i = 0;; ++i) { if (!uf.unite(edges[i][0] - 1, edges[i][1] - 1)) { return edges[i]; } } } }; -
class UnionFind: def __init__(self, n): self.p = list(range(n)) self.n = n def union(self, a, b): if self.find(a) == self.find(b): return False self.p[self.find(a)] = self.find(b) self.n -= 1 return True def find(self, x): if self.p[x] != x: self.p[x] = self.find(self.p[x]) return self.p[x] class Solution: def findRedundantDirectedConnection(self, edges: List[List[int]]) -> List[int]: n = len(edges) p = list(range(n + 1)) uf = UnionFind(n + 1) conflict = cycle = None for i, (u, v) in enumerate(edges): if p[v] != v: conflict = i else: p[v] = u if not uf.union(u, v): cycle = i if conflict is None: return edges[cycle] v = edges[conflict][1] if cycle is not None: return [p[v], v] return edges[conflict] # Solution 2 class UnionFind: __slots__ = "p", "size" def __init__(self, n: int): self.p: List[int] = list(range(n)) self.size: List[int] = [1] * n def find(self, x: int) -> int: if self.p[x] != x: self.p[x] = self.find(self.p[x]) return self.p[x] def union(self, a: int, b: int) -> bool: pa, pb = self.find(a), self.find(b) 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] return True class Solution: def findRedundantDirectedConnection(self, edges: List[List[int]]) -> List[int]: n = len(edges) ind = [0] * n for _, v in edges: ind[v - 1] += 1 dup = [i for i, (_, v) in enumerate(edges) if ind[v - 1] == 2] uf = UnionFind(n) if dup: for i, (u, v) in enumerate(edges): if i == dup[1]: continue if not uf.union(u - 1, v - 1): return edges[dup[0]] return edges[dup[1]] for i, (u, v) in enumerate(edges): if not uf.union(u - 1, v - 1): return edges[i] -
type unionFind struct { p []int n int } func newUnionFind(n int) *unionFind { p := make([]int, n) for i := range p { p[i] = i } return &unionFind{p, 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 { if uf.find(a) == uf.find(b) { return false } uf.p[uf.find(a)] = uf.find(b) uf.n-- return true } func findRedundantDirectedConnection(edges [][]int) []int { n := len(edges) p := make([]int, n+1) for i := range p { p[i] = i } uf := newUnionFind(n + 1) conflict, cycle := -1, -1 for i, e := range edges { u, v := e[0], e[1] if p[v] != v { conflict = i } else { p[v] = u if !uf.union(u, v) { cycle = i } } } if conflict == -1 { return edges[cycle] } v := edges[conflict][1] if cycle != -1 { return []int{p[v], v} } return edges[conflict] } // Solution 2 type unionFind struct { p, size []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} } 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), uf.find(b) 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] } return true } func findRedundantDirectedConnection(edges [][]int) []int { n := len(edges) ind := make([]int, n) for _, e := range edges { ind[e[1]-1]++ } dup := []int{} for i, e := range edges { if ind[e[1]-1] == 2 { dup = append(dup, i) } } uf := newUnionFind(n) if len(dup) > 0 { for i, e := range edges { if i == dup[1] { continue } if !uf.union(e[0]-1, e[1]-1) { return edges[dup[0]] } } return edges[dup[1]] } for _, e := range edges { if !uf.union(e[0]-1, e[1]-1) { return e } } return nil } -
function findRedundantDirectedConnection(edges: number[][]): number[] { const n = edges.length; const ind: number[] = Array(n).fill(0); for (const [_, v] of edges) { ++ind[v - 1]; } const dup: number[] = []; for (let i = 0; i < n; ++i) { if (ind[edges[i][1] - 1] === 2) { dup.push(i); } } const p: number[] = Array.from({ length: n }, (_, i) => i); const find = (x: number): number => { if (p[x] !== x) { p[x] = find(p[x]); } return p[x]; }; if (dup.length) { for (let i = 0; i < n; ++i) { if (i === dup[1]) { continue; } const [pu, pv] = [find(edges[i][0] - 1), find(edges[i][1] - 1)]; if (pu === pv) { return edges[dup[0]]; } p[pu] = pv; } return edges[dup[1]]; } for (let i = 0; ; ++i) { const [pu, pv] = [find(edges[i][0] - 1), find(edges[i][1] - 1)]; if (pu === pv) { return edges[i]; } p[pu] = pv; } } // Solution 2 class UnionFind { p: number[]; size: number[]; constructor(n: number) { this.p = Array.from({ length: n }, (_, i) => i); this.size = Array(n).fill(1); } find(x: number): number { if (this.p[x] !== x) { this.p[x] = this.find(this.p[x]); } return this.p[x]; } union(a: number, b: number): boolean { const [pa, pb] = [this.find(a), this.find(b)]; if (pa === pb) { return false; } if (this.size[pa] > this.size[pb]) { this.p[pb] = pa; this.size[pa] += this.size[pb]; } else { this.p[pa] = pb; this.size[pb] += this.size[pa]; } return true; } } function findRedundantDirectedConnection(edges: number[][]): number[] { const n = edges.length; const ind: number[] = Array(n).fill(0); for (const [_, v] of edges) { ++ind[v - 1]; } const dup: number[] = []; for (let i = 0; i < n; ++i) { if (ind[edges[i][1] - 1] === 2) { dup.push(i); } } const uf = new UnionFind(n); if (dup.length) { for (let i = 0; i < n; ++i) { if (i === dup[1]) { continue; } if (!uf.union(edges[i][0] - 1, edges[i][1] - 1)) { return edges[dup[0]]; } } return edges[dup[1]]; } for (let i = 0; ; ++i) { if (!uf.union(edges[i][0] - 1, edges[i][1] - 1)) { return edges[i]; } } } -
/** * @param {number[][]} edges * @return {number[]} */ var findRedundantDirectedConnection = function (edges) { const n = edges.length; const ind = Array(n).fill(0); for (const [_, v] of edges) { ++ind[v - 1]; } const dup = []; for (let i = 0; i < n; ++i) { if (ind[edges[i][1] - 1] === 2) { dup.push(i); } } const p = Array.from({ length: n }, (_, i) => i); const find = x => { if (p[x] !== x) { p[x] = find(p[x]); } return p[x]; }; if (dup.length) { for (let i = 0; i < n; ++i) { if (i === dup[1]) { continue; } const [pu, pv] = [find(edges[i][0] - 1), find(edges[i][1] - 1)]; if (pu === pv) { return edges[dup[0]]; } p[pu] = pv; } return edges[dup[1]]; } for (let i = 0; ; ++i) { const [pu, pv] = [find(edges[i][0] - 1), find(edges[i][1] - 1)]; if (pu === pv) { return edges[i]; } p[pu] = pv; } }; // Solution 2 class UnionFind { constructor(n) { this.p = Array.from({ length: n }, (_, i) => i); this.size = Array(n).fill(1); } find(x) { if (this.p[x] !== x) { this.p[x] = this.find(this.p[x]); } return this.p[x]; } union(a, b) { const pa = this.find(a); const pb = this.find(b); if (pa === pb) { return false; } if (this.size[pa] > this.size[pb]) { this.p[pb] = pa; this.size[pa] += this.size[pb]; } else { this.p[pa] = pb; this.size[pb] += this.size[pa]; } return true; } } /** * @param {number[][]} edges * @return {number[]} */ var findRedundantDirectedConnection = function (edges) { const n = edges.length; const ind = Array(n).fill(0); for (const [_, v] of edges) { ++ind[v - 1]; } const dup = []; for (let i = 0; i < n; ++i) { if (ind[edges[i][1] - 1] === 2) { dup.push(i); } } const uf = new UnionFind(n); if (dup.length) { for (let i = 0; i < n; ++i) { if (i === dup[1]) { continue; } if (!uf.union(edges[i][0] - 1, edges[i][1] - 1)) { return edges[dup[0]]; } } return edges[dup[1]]; } for (let i = 0; ; ++i) { if (!uf.union(edges[i][0] - 1, edges[i][1] - 1)) { return edges[i]; } } };