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// 2493. Divide Nodes Into the Maximum Number of Groups
#include "leetcode.h"
/*
* you are given a positive integer 'n' representing the number of nodes in an
* undirected graph. the nodes are labeled from 1 to 'n'. you are also given a
* 2d integer array 'edges', where 'edges[i] = [a_i, b_i]' indicates that there
* is a bidirectional edge between nodes 'a_i' and 'b_i'. divide the nodes of
* the graph into 'm' groups such that each node in the graph belongs to exactly
* one group, and for every pair of nodes in the graph that are connected by an
* edge '[a_i, b_i]', then '|y - x| = 1'. return the maximum number of groups
* into which you can divide the nodes. return -1 if it is impossible.
*/
class Solution {
public:
int magnificentSets(int n, vector<vector<int>> &edges) {
vector<int> vis(n);
vector<vector<int>> graph(n), group;
for (auto e : edges) {
graph[e[0] - 1].push_back(e[1] - 1);
graph[e[1] - 1].push_back(e[0] - 1);
}
for (int i = 0; i < n; ++i)
if (!vis[i]) {
vis[i] = true;
stack<int> stk;
stk.push(i);
group.push_back({i});
while (stk.size()) {
auto u = stk.top();
stk.pop();
for (auto &v : graph[u])
if (!vis[v]) {
vis[v] = vis[u] + 1;
stk.push(v);
group.back().push_back(v);
} else if ((vis[u] & 1) == (vis[v] & 1))
return -1;
}
}
auto bfs = [&](int x) {
int res = 0;
vector<bool> vis(n);
vis[x] = true;
queue<int> q;
q.push(x);
for (; q.size(); ++res)
for (int size = q.size(); size; --size) {
auto u = q.front();
q.pop();
for (auto &v : graph[u])
if (!vis[v]) {
vis[v] = true;
q.push(v);
}
}
return res;
};
int ans = 0;
for (auto &g : group) {
transform(g.begin(), g.end(), g.begin(), bfs);
ans += *max_element(g.begin(), g.end());
}
return ans;
}
};
int main() {
Solution obj;
vvd e1 = {{1, 2}, {1, 4}, {1, 5}, {2, 6}, {2, 3}, {4, 6}};
vvd e2 = {{1, 2}, {2, 3}, {3, 1}};
printf("%d\n", obj.magnificentSets(6, e1)); // expect: 4
printf("%d\n", obj.magnificentSets(3, e2)); // expect: -1
}
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