# 1743. Restore the Array From Adjacent Pairs

## Description

There is an integer array nums that consists of n unique elements, but you have forgotten it. However, you do remember every pair of adjacent elements in nums.

You are given a 2D integer array adjacentPairs of size n - 1 where each adjacentPairs[i] = [ui, vi] indicates that the elements ui and vi are adjacent in nums.

It is guaranteed that every adjacent pair of elements nums[i] and nums[i+1] will exist in adjacentPairs, either as [nums[i], nums[i+1]] or [nums[i+1], nums[i]]. The pairs can appear in any order.

Return the original array nums. If there are multiple solutions, return any of them.

Example 1:

Input: adjacentPairs = [[2,1],[3,4],[3,2]]
Output: [1,2,3,4]
Notice that adjacentPairs[i] may not be in left-to-right order.


Example 2:

Input: adjacentPairs = [[4,-2],[1,4],[-3,1]]
Output: [-2,4,1,-3]
Explanation: There can be negative numbers.
Another solution is [-3,1,4,-2], which would also be accepted.


Example 3:

Input: adjacentPairs = [[100000,-100000]]
Output: [100000,-100000]


Constraints:

• nums.length == n
• adjacentPairs.length == n - 1
• adjacentPairs[i].length == 2
• 2 <= n <= 105
• -105 <= nums[i], ui, vi <= 105
• There exists some nums that has adjacentPairs as its pairs.

## Solutions

Traverse the graph from the point where the degree is one.

• class Solution {
int n = adjacentPairs.length + 1;
Map<Integer, List<Integer>> g = new HashMap<>();
for (int[] e : adjacentPairs) {
int a = e[0], b = e[1];
}
int[] ans = new int[n];
for (Map.Entry<Integer, List<Integer>> entry : g.entrySet()) {
if (entry.getValue().size() == 1) {
ans[0] = entry.getKey();
ans[1] = entry.getValue().get(0);
break;
}
}
for (int i = 2; i < n; ++i) {
List<Integer> v = g.get(ans[i - 1]);
ans[i] = v.get(1) == ans[i - 2] ? v.get(0) : v.get(1);
}
return ans;
}
}

• class Solution {
public:
int n = adjacentPairs.size() + 1;
unordered_map<int, vector<int>> g;
for (auto& e : adjacentPairs) {
int a = e[0], b = e[1];
g[a].push_back(b);
g[b].push_back(a);
}
vector<int> ans(n);
for (auto& [k, v] : g) {
if (v.size() == 1) {
ans[0] = k;
ans[1] = v[0];
break;
}
}
for (int i = 2; i < n; ++i) {
auto v = g[ans[i - 1]];
ans[i] = v[0] == ans[i - 2] ? v[1] : v[0];
}
return ans;
}
};

• class Solution:
def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
g = defaultdict(list)
g[a].append(b)
g[b].append(a)
ans = [0] * n
for i, v in g.items():
if len(v) == 1:
ans[0] = i
ans[1] = v[0]
break
for i in range(2, n):
v = g[ans[i - 1]]
ans[i] = v[0] if v[1] == ans[i - 2] else v[1]
return ans


• func restoreArray(adjacentPairs [][]int) []int {
g := map[int][]int{}
for _, e := range adjacentPairs {
a, b := e[0], e[1]
g[a] = append(g[a], b)
g[b] = append(g[b], a)
}
ans := make([]int, n)
for k, v := range g {
if len(v) == 1 {
ans[0] = k
ans[1] = v[0]
break
}
}
for i := 2; i < n; i++ {
v := g[ans[i-1]]
ans[i] = v[0]
if v[0] == ans[i-2] {
ans[i] = v[1]
}
}
return ans
}

• public class Solution {
int n = adjacentPairs.Length + 1;
Dictionary<int, List<int>> g = new Dictionary<int, List<int>>();

foreach (int[] e in adjacentPairs) {
int a = e[0], b = e[1];
if (!g.ContainsKey(a)) {
g[a] = new List<int>();
}
if (!g.ContainsKey(b)) {
g[b] = new List<int>();
}
}

int[] ans = new int[n];

foreach (var entry in g) {
if (entry.Value.Count == 1) {
ans[0] = entry.Key;
ans[1] = entry.Value[0];
break;
}
}

for (int i = 2; i < n; ++i) {
List<int> v = g[ans[i - 1]];
ans[i] = v[1] == ans[i - 2] ? v[0] : v[1];
}

return ans;
}
}