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 <= 105

• -105 <= nums[i], ui, vi <= 105

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

Solution

/**
 * @param {number[][]} adjacentPairs
 * @return {number[]}
 */
var restoreArray = function(adjacentPairs) {
    var map = {};
    for (var i = 0 ; i < adjacentPairs.length; i++) {
        map[adjacentPairs[i][0]] = map[adjacentPairs[i][0]] || [];
        map[adjacentPairs[i][1]] = map[adjacentPairs[i][1]] || [];
        map[adjacentPairs[i][0]].push(adjacentPairs[i][1]);
        map[adjacentPairs[i][1]].push(adjacentPairs[i][0]);
    }
    var root = Number(Object.keys(map).find(num => map[num].length === 1));
    var res = [root];
    var last = root;
    var now = map[root][0];
    while (now !== undefined) {
        var next = map[now][0] === last ? map[now][1] : map[now][0];
        res.push(now);
        last = now;
        now = next;
    }
    return res;
};

Explain:

nope.

Complexity:

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