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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/valid_arrangement_pairs.cpp | |
parent | 5d7cdc215e4ba1ec4cbb24c084cccb5e5e641468 (diff) | |
download | leetcode-7c60dee83b436e8e8a667bf9f191beced8de4fc3.tar.gz leetcode-7c60dee83b436e8e8a667bf9f191beced8de4fc3.zip |
chore: move cpp and rust code to directory
Diffstat (limited to 'other/valid_arrangement_pairs.cpp')
-rw-r--r-- | other/valid_arrangement_pairs.cpp | 73 |
1 files changed, 73 insertions, 0 deletions
diff --git a/other/valid_arrangement_pairs.cpp b/other/valid_arrangement_pairs.cpp new file mode 100644 index 0000000..0493141 --- /dev/null +++ b/other/valid_arrangement_pairs.cpp @@ -0,0 +1,73 @@ +// 2097. Valid Arrangement of Pairs +#include "leetcode.h" + +/* + * you are given a 0-indexed 2d integer array 'pairs' where 'pairs[i] = + * [start_i, end_i]'. an arrangement of 'pairs' is valid for every index 'i' + * where '1 <= i < pairs.length', we have 'end_i-1 == start_i'. return any valid + * arrangement of 'pairs'. note, the inputs will be generated such that there + * exists a valid arrangement of 'pairs'. + */ + +class Solution { +private: + void euler(unordered_map<int, stack<int>> &adj, vector<vector<int>> &res, + int curr) { + auto &st = adj[curr]; + while (!st.empty()) { + int n = st.top(); + st.pop(); + euler(adj, res, n); + res.push_back({curr, n}); + } + } + +public: + vector<vector<int>> validArrangement(vector<vector<int>> &pairs) { + int m = pairs.size(); + unordered_map<int, stack<int>> adj; + unordered_map<int, int> in, out; + adj.reserve(m); + in.reserve(m); + out.reserve(m); + for (int i = 0; i < m; i++) { + int u = pairs[i][0], v = pairs[i][1]; + in[v]++; + out[u]++; + adj[u].push(v); + } + int start = -1; + for (auto &p : adj) { + int i = p.first; + if (out[i] - in[i] == 1) + start = i; + } + if (start == -1) + start = adj.begin()->first; + vector<vector<int>> ans; + euler(adj, ans, start); + reverse(ans.begin(), ans.end()); + return ans; + } +}; + +int main() { + Solution obj; + vvd p1 = {{5, 1}, {4, 5}, {11, 9}, {9, 4}}, + p2 = {{1, 3}, {3, 2}, {2, 1}}, p3 = {{1, 2}, {1, 3}, {2, 1}}; + vvd va1 = obj.validArrangement(p1); + vvd va2 = obj.validArrangement(p2); + vvd va3 = obj.validArrangement(p3); + for (int i = 0; i < va1.size(); i++) + for (int j = 0; j < va1[0].size(); j++) + printf("%d ", va1[i][j]); // expect: [[11,9],[9,4],[4,5],[5,1]] + printf("\n"); + for (int i = 0; i < va2.size(); i++) + for (int j = 0; j < va2[0].size(); j++) + printf("%d ", va2[i][j]); // expect: [[1,3],[3,2],[2,1]] + printf("\n"); + for (int i = 0; i < va3.size(); i++) + for (int j = 0; j < va3[0].size(); j++) + printf("%d ", va1[i][j]); // expect: [[1,2],[2,1],[1,3]] + printf("\n"); +} |