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1743. Restore the Array From Adjacent Pairs

Description

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.

Solutions

Solution 1

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class Solution:
    def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
        g = defaultdict(list)
        for a, b in adjacentPairs:
            g[a].append(b)
            g[b].append(a)
        n = len(adjacentPairs) + 1
        ans = [0] * n
        for i, v in g.items():
            if len(v) == 1:
                ans[0] = i
                ans[1] = v[0]
                break
        for i in range(2, n):
            v = g[ans[i - 1]]
            ans[i] = v[0] if v[1] == ans[i - 2] else v[1]
        return ans

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class Solution {
    public int[] restoreArray(int[][] adjacentPairs) {
        int n = adjacentPairs.length + 1;
        Map<Integer, List<Integer>> g = new HashMap<>();
        for (int[] e : adjacentPairs) {
            int a = e[0], b = e[1];
            g.computeIfAbsent(a, k -> new ArrayList<>()).add(b);
            g.computeIfAbsent(b, k -> new ArrayList<>()).add(a);
        }
        int[] ans = new int[n];
        for (Map.Entry<Integer, List<Integer>> entry : g.entrySet()) {
            if (entry.getValue().size() == 1) {
                ans[0] = entry.getKey();
                ans[1] = entry.getValue().get(0);
                break;
            }
        }
        for (int i = 2; i < n; ++i) {
            List<Integer> v = g.get(ans[i - 1]);
            ans[i] = v.get(1) == ans[i - 2] ? v.get(0) : v.get(1);
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> restoreArray(vector<vector<int>>& adjacentPairs) {
        int n = adjacentPairs.size() + 1;
        unordered_map<int, vector<int>> g;
        for (auto& e : adjacentPairs) {
            int a = e[0], b = e[1];
            g[a].push_back(b);
            g[b].push_back(a);
        }
        vector<int> ans(n);
        for (auto& [k, v] : g) {
            if (v.size() == 1) {
                ans[0] = k;
                ans[1] = v[0];
                break;
            }
        }
        for (int i = 2; i < n; ++i) {
            auto v = g[ans[i - 1]];
            ans[i] = v[0] == ans[i - 2] ? v[1] : v[0];
        }
        return ans;
    }
};

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func restoreArray(adjacentPairs [][]int) []int {
    n := len(adjacentPairs) + 1
    g := map[int][]int{}
    for _, e := range adjacentPairs {
        a, b := e[0], e[1]
        g[a] = append(g[a], b)
        g[b] = append(g[b], a)
    }
    ans := make([]int, n)
    for k, v := range g {
        if len(v) == 1 {
            ans[0] = k
            ans[1] = v[0]
            break
        }
    }
    for i := 2; i < n; i++ {
        v := g[ans[i-1]]
        ans[i] = v[0]
        if v[0] == ans[i-2] {
            ans[i] = v[1]
        }
    }
    return ans
}

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public class Solution {
    public int[] RestoreArray(int[][] adjacentPairs) {
        int n = adjacentPairs.Length + 1;
        Dictionary<int, List<int>> g = new Dictionary<int, List<int>>();

        foreach (int[] e in adjacentPairs) {
            int a = e[0], b = e[1];
            if (!g.ContainsKey(a)) {
                g[a] = new List<int>();
            }
            if (!g.ContainsKey(b)) {
                g[b] = new List<int>();
            }
            g[a].Add(b);
            g[b].Add(a);
        }

        int[] ans = new int[n];

        foreach (var entry in g) {
            if (entry.Value.Count == 1) {
                ans[0] = entry.Key;
                ans[1] = entry.Value[0];
                break;
            }
        }

        for (int i = 2; i < n; ++i) {
            List<int> v = g[ans[i - 1]];
            ans[i] = v[1] == ans[i - 2] ? v[0] : v[1];
        }

        return ans;
    }
}

Solution 2

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class Solution:
    def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
        def dfs(i, fa):
            ans.append(i)
            for j in g[i]:
                if j != fa:
                    dfs(j, i)

        g = defaultdict(list)
        for a, b in adjacentPairs:
            g[a].append(b)
            g[b].append(a)
        i = next(i for i, v in g.items() if len(v) == 1)
        ans = []
        dfs(i, 1e6)
        return ans

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class Solution {
    private Map<Integer, List<Integer>> g = new HashMap<>();
    private int[] ans;

    public int[] restoreArray(int[][] adjacentPairs) {
        for (var e : adjacentPairs) {
            int a = e[0], b = e[1];
            g.computeIfAbsent(a, k -> new ArrayList<>()).add(b);
            g.computeIfAbsent(b, k -> new ArrayList<>()).add(a);
        }
        int n = adjacentPairs.length + 1;
        ans = new int[n];
        for (var e : g.entrySet()) {
            if (e.getValue().size() == 1) {
                dfs(e.getKey(), 1000000, 0);
                break;
            }
        }
        return ans;
    }

    private void dfs(int i, int fa, int k) {
        ans[k++] = i;
        for (int j : g.get(i)) {
            if (j != fa) {
                dfs(j, i, k);
            }
        }
    }
}

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class Solution {
public:
    vector<int> restoreArray(vector<vector<int>>& adjacentPairs) {
        unordered_map<int, vector<int>> g;
        for (auto& e : adjacentPairs) {
            int a = e[0], b = e[1];
            g[a].emplace_back(b);
            g[b].emplace_back(a);
        }
        int n = adjacentPairs.size() + 1;
        vector<int> ans;
        function<void(int, int)> dfs = [&](int i, int fa) {
            ans.emplace_back(i);
            for (int& j : g[i]) {
                if (j != fa) {
                    dfs(j, i);
                }
            }
        };
        for (auto& [i, v] : g) {
            if (v.size() == 1) {
                dfs(i, 1e6);
                break;
            }
        }
        return ans;
    }
};

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func restoreArray(adjacentPairs [][]int) []int {
    g := map[int][]int{}
    for _, e := range adjacentPairs {
        a, b := e[0], e[1]
        g[a] = append(g[a], b)
        g[b] = append(g[b], a)
    }
    ans := []int{}
    var dfs func(i, fa int)
    dfs = func(i, fa int) {
        ans = append(ans, i)
        for _, j := range g[i] {
            if j != fa {
                dfs(j, i)
            }
        }
    }
    for i, v := range g {
        if len(v) == 1 {
            dfs(i, 1000000)
            break
        }
    }
    return ans
}

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