1743

1743. Restore the Array From Adjacent Pairs (Medium)

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.

Companies: Uber, Capital One, Microsoft, Robinhood

Related Topics:
Array, Hash Table

Hints:

• Find the first element of nums - it will only appear once in adjacentPairs.
• The adjacent pairs are like edges of a graph. Perform a depth-first search from the first element.

Solution 1.

// OJ: https://leetcode.com/problems/restore-the-array-from-adjacent-pairs
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> restoreArray(vector<vector<int>>& A) {
        int N = A.size() + 1, first = -1;
        unordered_map<int, vector<int>> G;
        for (auto &p : A) { // build graph
            int u = p[0], v = p[1];
            G[u].push_back(v);
            G[v].push_back(u);
        }
        vector<int> ans;
        for (auto &[n, v] : G) { // find an edge node
            if (v.size() == 1) {
                ans.push_back(n);
                ans.push_back(v[0]);
                break;
            }
        }
        while (ans.size() < N) { // keep finding the next node
            int n = ans.back(), prev = ans[ans.size() - 2];
            if (G[n][0] != prev) ans.push_back(G[n][0]);
            else ans.push_back(G[n][1]);
        }
        return ans;
    }
};