1743. Restore the Array From Adjacent Pairs

Problem

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]
Explanation: This array has all its adjacent pairs in adjacentPairs.
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 <= 10^5

• -10^5 <= nums[i], ui, vi <= 10^5

• There exists some nums that has adjacentPairs as its pairs.

Solution (Java)

class Solution {
    public int[] restoreArray(int[][] adjacentPairs) {
        if (adjacentPairs == null || adjacentPairs.length == 0) {
            return new int[0];
        }
        if (adjacentPairs.length == 1) {
            return adjacentPairs[0];
        }
        Map<Integer, List<Integer>> graph = new HashMap<>();
        for (int[] pair : adjacentPairs) {
            graph.computeIfAbsent(pair[0], k -> new ArrayList<>()).add(pair[1]);
            graph.computeIfAbsent(pair[1], k -> new ArrayList<>()).add(pair[0]);
        }
        int[] res = new int[graph.size()];
        for (Map.Entry<Integer, List<Integer>> entry : graph.entrySet()) {
            if (entry.getValue().size() == 1) {
                res[0] = entry.getKey();
                break;
            }
        }
        res[1] = graph.get(res[0]).get(0);
        for (int i = 2; i < res.length; i++) {
            for (int cur : graph.get(res[i - 1])) {
                if (cur != res[i - 2]) {
                    res[i] = cur;
                    break;
                }
            }
        }
        return res;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).