Restore the Array From Adjacent Pairs

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

SOLUTION:

class Solution:
    def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
        nums = {}
        singles = set()
        for a, b in adjacentPairs:
            nums[a] = nums.get(a, []) + [b]
            nums[b] = nums.get(b, []) + [a]
            if a in singles:
                singles.remove(a)
            else:
                singles.add(a)
            if b in singles:
                singles.remove(b)
            else:
                singles.add(b)
        first = list(singles)[0]
        n = len(nums)
        op = []
        op.extend([first, nums[first][0]])
        while len(op) < n:
            prev = op[-1]
            prevprev = op[-2]
            op.extend([k for k in nums[prev] if k != prevprev])
        return op