In this guide, you will get 685. Redundant Connection II LeetCode Solution with the best time and space complexity. The solution to Redundant Connection II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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.
n == edges.length
3 <= n <= 1000
edges[i].length == 2
1 <= ui, vi <= n
ui != vi
Complexity Analysis
Time Complexity: O(|V|)
Space Complexity: O(|V|)
685. Redundant Connection II LeetCode Solution in C++
class UnionFind {
public:
UnionFind(int n) : id(n), rank(n) {
iota(id.begin(), id.end(), 0);
}
bool unionByRank(int u, int v) {
const int i = find(u);
const int j = find(v);
if (i == j)
return false;
if (rank[i] < rank[j]) {
id[i] = j;
} else if (rank[i] > rank[j]) {
id[j] = i;
} else {
id[i] = j;
++rank[j];
}
return true;
}
private:
vector<int> id;
vector<int> rank;
int find(int u) {
return id[u] == u ? u : id[u] = find(id[u]);
}
};
class Solution {
public:
vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {
vector<int> ids(edges.size() + 1);
int nodeWithTwoParents = 0;
for (const vector<int>& edge : edges) {
const int v = edge[1];
if (++ids[v] == 2) {
nodeWithTwoParents = v;
break;
}
}
// If there is no edge with two ids, don't skip any edge.
if (nodeWithTwoParents == 0)
return findRedundantDirectedConnection(edges, -1);
for (int i = edges.size() - 1; i >= 0; --i)
if (edges[i][1] == nodeWithTwoParents)
// Try to delete the edges[i].
if (findRedundantDirectedConnection(edges, i).empty())
return edges[i];
throw;
}
vector<int> findRedundantDirectedConnection(const vector<vector<int>>& edges,
int skippedEdgeIndex) {
UnionFind uf(edges.size() + 1);
for (int i = 0; i < edges.size(); ++i) {
if (i == skippedEdgeIndex)
continue;
if (!uf.unionByRank(edges[i][0], edges[i][1]))
return edges[i];
}
return {};
}
}; /* code provided by PROGIEZ */
685. Redundant Connection II LeetCode Solution in Java
class UnionFind {
public UnionFind(int n) {
id = new int[n];
rank = new int[n];
for (int i = 0; i < n; ++i)
id[i] = i;
}
public boolean unionByRank(int u, int v) {
final int i = find(u);
final int j = find(v);
if (i == j)
return false;
if (rank[i] < rank[j]) {
id[i] = j;
} else if (rank[i] > rank[j]) {
id[j] = i;
} else {
id[i] = j;
++rank[j];
}
return true;
}
private int[] id;
private int[] rank;
private int find(int u) {
return id[u] == u ? u : (id[u] = find(id[u]));
}
}
class Solution {
public int[] findRedundantDirectedConnection(int[][] edges) {
int[] ids = new int[edges.length + 1];
int nodeWithTwoParents = 0;
for (int[] edge : edges) {
final int v = edge[1];
if (++ids[v] == 2) {
nodeWithTwoParents = v;
break;
}
}
// If there is no edge with two ids, don't skip any edge.
if (nodeWithTwoParents == 0)
return findRedundantDirectedConnection(edges, -1);
for (int i = edges.length - 1; i >= 0; --i)
if (edges[i][1] == nodeWithTwoParents)
// Try to delete the edges[i].
if (findRedundantDirectedConnection(edges, i).length == 0)
return edges[i];
throw new IllegalArgumentException();
}
private int[] findRedundantDirectedConnection(int[][] edges, int skippedEdgeIndex) {
UnionFind uf = new UnionFind(edges.length + 1);
for (int i = 0; i < edges.length; ++i) {
if (i == skippedEdgeIndex)
continue;
if (!uf.unionByRank(edges[i][0], edges[i][1]))
return edges[i];
}
return new int[] {};
}
} // code provided by PROGIEZ
685. Redundant Connection II LeetCode Solution in Python
class UnionFind:
def __init__(self, n: int):
self.id = list(range(n))
self.rank = [0] * n
def unionByRank(self, u: int, v: int) -> bool:
i = self._find(u)
j = self._find(v)
if i == j:
return False
if self.rank[i] < self.rank[j]:
self.id[i] = j
elif self.rank[i] > self.rank[j]:
self.id[j] = i
else:
self.id[i] = j
self.rank[j] += 1
return True
def _find(self, u: int) -> int:
if self.id[u] != u:
self.id[u] = self._find(self.id[u])
return self.id[u]
class Solution:
def findRedundantDirectedConnection(
self, edges: list[list[int]],
) -> list[int]:
ids = [0] * (len(edges) + 1)
nodeWithTwoParents = 0
for _, v in edges:
ids[v] += 1
if ids[v] == 2:
nodeWithTwoParents = v
def findRedundantDirectedConnection(skippedEdgeIndex: int) -> list[int]:
uf = UnionFind(len(edges) + 1)
for i, edge in enumerate(edges):
if i == skippedEdgeIndex:
continue
if not uf.unionByRank(edge[0], edge[1]):
return edge
return []
# If there is no edge with two ids, don't skip any edge.
if nodeWithTwoParents == 0:
return findRedundantDirectedConnection(-1)
for i in reversed(range(len(edges))):
_, v = edges[i]
if v == nodeWithTwoParents:
# Try to delete the edges[i].
if not findRedundantDirectedConnection(i):
return edges[i] # code by PROGIEZ
