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-rw-r--r--valid_arrangement_pairs.cpp73
-rw-r--r--valid_arrangement_pairs.py44
2 files changed, 117 insertions, 0 deletions
diff --git a/valid_arrangement_pairs.cpp b/valid_arrangement_pairs.cpp
new file mode 100644
index 0000000..f1f8b20
--- /dev/null
+++ b/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(int) p1 = {{5, 1}, {4, 5}, {11, 9}, {9, 4}},
+ p2 = {{1, 3}, {3, 2}, {2, 1}}, p3 = {{1, 2}, {1, 3}, {2, 1}};
+ vvd(int) va1 = obj.validArrangement(p1);
+ vvd(int) va2 = obj.validArrangement(p2);
+ vvd(int) 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");
+}
diff --git a/valid_arrangement_pairs.py b/valid_arrangement_pairs.py
new file mode 100644
index 0000000..528c883
--- /dev/null
+++ b/valid_arrangement_pairs.py
@@ -0,0 +1,44 @@
+# 2097. Valid Arrangement of Pairs
+from collections import defaultdict
+
+"""
+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(object):
+ def validArrangement(self, pairs):
+ """
+ :type pairs: List[List[int]]
+ :rtype: List[List[int]]
+ """
+ graph, degree = defaultdict(list), defaultdict(int)
+ for x, y in pairs:
+ graph[x].append(y)
+ degree[x] += 1
+ degree[y] -= 1
+ for k in degree:
+ if degree[k] == 1:
+ x = k
+ break
+ ans = []
+
+ def euler(x):
+ while graph[x]:
+ euler(graph[x].pop())
+ ans.append(x)
+
+ euler(x)
+ ans.reverse()
+ return [[ans[i], ans[i + 1]] for i in range(len(ans) - 1)]
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.validArrangement(pairs=[[5, 1], [4, 5], [11, 9], [9, 4]]))
+ print(obj.validArrangement(pairs=[[1, 3], [3, 2], [2, 1]]))
+ print(obj.validArrangement(pairs=[[1, 2], [1, 3], [2, 1]]))