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author | Jack Sangdahl <[email protected]> | 2025-03-10 00:36:13 -0600 |
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committer | Jack Sangdahl <[email protected]> | 2025-03-10 00:36:13 -0600 |
commit | 7c60dee83b436e8e8a667bf9f191beced8de4fc3 (patch) | |
tree | eb7a24ee2067f92245c57bb118e85a1baf15df50 /other/divide_nodes_max_groups.cpp | |
parent | 5d7cdc215e4ba1ec4cbb24c084cccb5e5e641468 (diff) | |
download | leetcode-7c60dee83b436e8e8a667bf9f191beced8de4fc3.tar.gz leetcode-7c60dee83b436e8e8a667bf9f191beced8de4fc3.zip |
chore: move cpp and rust code to directory
Diffstat (limited to 'other/divide_nodes_max_groups.cpp')
-rw-r--r-- | other/divide_nodes_max_groups.cpp | 75 |
1 files changed, 75 insertions, 0 deletions
diff --git a/other/divide_nodes_max_groups.cpp b/other/divide_nodes_max_groups.cpp new file mode 100644 index 0000000..ea07a34 --- /dev/null +++ b/other/divide_nodes_max_groups.cpp @@ -0,0 +1,75 @@ +// 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 +} |