1579. Remove Max Number of Edges to Keep Graph Fully Traversable LeetCode Solution
In this guide, you will get 1579. Remove Max Number of Edges to Keep Graph Fully Traversable LeetCode Solution with the best time and space complexity. The solution to Remove Max Number of Edges to Keep Graph Fully Traversable 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.
Table of Contents
- Problem Statement
- Complexity Analysis
- Remove Max Number of Edges to Keep Graph Fully Traversable solution in C++
- Remove Max Number of Edges to Keep Graph Fully Traversable solution in Java
- Remove Max Number of Edges to Keep Graph Fully Traversable solution in Python
- Additional Resources
Problem Statement of Remove Max Number of Edges to Keep Graph Fully Traversable
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.
Complexity Analysis
- Time Complexity: O(\texttt{sort})
- Space Complexity: O(n)
1579. Remove Max Number of Edges to Keep Graph Fully Traversable LeetCode Solution in C++
class UnionFind {
public:
UnionFind(int n) : count(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];
}
--count;
return true;
}
int getCount() const {
return count;
}
private:
int count;
vector<int> id;
vector<int> rank;
int find(int u) {
return id[u] == u ? u : id[u] = find(id[u]);
}
};
class Solution {
public:
int maxNumEdgesToRemove(int n, vector<vector<int>>& edges) {
UnionFind alice(n);
UnionFind bob(n);
int requiredEdges = 0;
// Greedily put type 3 edges in the front.
ranges::sort(edges, ranges::greater{},
[](const vector<int>& edge) { return edge[0]; });
for (const vector<int>& edge : edges) {
const int type = edge[0];
const int u = edge[1] - 1;
const int v = edge[2] - 1;
switch (type) {
case 3: // Can be traversed by Alice and Bob.
// Note that we should use | instead of || because if the first
// expression is true, short-circuiting will skip the second
// expression.
if (alice.unionByRank(u, v) | bob.unionByRank(u, v))
++requiredEdges;
break;
case 2: // Can be traversed by Bob.
if (bob.unionByRank(u, v))
++requiredEdges;
break;
case 1: // Can be traversed by Alice.
if (alice.unionByRank(u, v))
++requiredEdges;
break;
}
}
return alice.getCount() == 1 && bob.getCount() == 1
? edges.size() - requiredEdges
: -1;
}
};
/* code provided by PROGIEZ */1579. Remove Max Number of Edges to Keep Graph Fully Traversable LeetCode Solution in Java
class UnionFind {
public UnionFind(int n) {
count = 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];
}
--count;
return true;
}
public int getCount() {
return count;
}
private int count;
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 maxNumEdgesToRemove(int n, int[][] edges) {
UnionFind alice = new UnionFind(n);
UnionFind bob = new UnionFind(n);
int requiredEdges = 0;
// Greedily put type 3 edges in the front.
Arrays.sort(edges, (a, b) -> Integer.compare(b[0], a[0]));
for (int[] edge : edges) {
final int type = edge[0];
final int u = edge[1] - 1;
final int v = edge[2] - 1;
switch (type) {
case 3: // Can be traversed by Alice and Bob.
// Note that we should use | instead of || because if the first
// expression is true, short-circuiting will skip the second
// expression.
if (alice.unionByRank(u, v) | bob.unionByRank(u, v))
++requiredEdges;
break;
case 2: // Can be traversed by Bob.
if (bob.unionByRank(u, v))
++requiredEdges;
break;
case 1: // Can be traversed by Alice.
if (alice.unionByRank(u, v))
++requiredEdges;
}
}
return alice.getCount() == 1 && bob.getCount() == 1 ? edges.length - requiredEdges : -1;
}
}
// code provided by PROGIEZ1579. Remove Max Number of Edges to Keep Graph Fully Traversable LeetCode Solution in Python
class UnionFind:
def __init__(self, n: int):
self.count = n
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
self.count -= 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 maxNumEdgesToRemove(self, n: int, edges: list[list[int]]) -> int:
alice = UnionFind(n)
bob = UnionFind(n)
requiredEdges = 0
# Greedily put type 3 edges in the front.
for type_, u, v in sorted(edges, reverse=True):
u -= 1
v -= 1
if type_ == 3: # Can be traversed by Alice and Bob.
# Note that we should use | instead of or because if the first
# expression is True, short-circuiting will skip the second
# expression.
if alice.unionByRank(u, v) | bob.unionByRank(u, v):
requiredEdges += 1
elif type_ == 2: # Can be traversed by Bob.
if bob.unionByRank(u, v):
requiredEdges += 1
else: # type == 1 Can be traversed by Alice.
if alice.unionByRank(u, v):
requiredEdges += 1
return (len(edges) - requiredEdges
if alice.count == 1 and bob.count == 1
else -1)
# code by PROGIEZAdditional Resources
- Explore all LeetCode problem solutions at Progiez here
- Explore all problems on LeetCode website here
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