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// 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");
}
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