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authorJack Sangdahl <[email protected]>2025-03-10 00:36:13 -0600
committerJack Sangdahl <[email protected]>2025-03-10 00:36:13 -0600
commit7c60dee83b436e8e8a667bf9f191beced8de4fc3 (patch)
treeeb7a24ee2067f92245c57bb118e85a1baf15df50 /other/divide_nodes_max_groups.cpp
parent5d7cdc215e4ba1ec4cbb24c084cccb5e5e641468 (diff)
downloadleetcode-7c60dee83b436e8e8a667bf9f191beced8de4fc3.tar.gz
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chore: move cpp and rust code to directory
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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
+}