1743. Restore the Array From Adjacent Pairs

Problem Link

Description

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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

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Python3

class Solution:
    def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
        M = len(adjacentPairs)
        graph = defaultdict(list)
 
        for a, b in adjacentPairs:
            graph[a].append(b)
            graph[b].append(a)
        
        res = []
        for k in graph.keys():
            if len(graph[k]) == 1:
                res.append(k)
                res.append(graph[k][0])
                break
        
        while len(res) < M + 1:
            prev, head = res[-2], res[-1]
 
            adj = graph[head]
 
            if adj[0] == prev:
                res.append(adj[1])
            else:
                res.append(adj[0])
        
        return res