LeetCode in Kotlin

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.

Solution

class Solution {
    fun restoreArray(adjacentPairs: Array<IntArray>): IntArray {
        if (adjacentPairs.isEmpty()) {
            return IntArray(0)
        }
        if (adjacentPairs.size == 1) {
            return adjacentPairs[0]
        }
        val graph: MutableMap<Int, MutableList<Int>> = HashMap()
        for (pair: IntArray in adjacentPairs) {
            graph.computeIfAbsent(pair[0]) { _: Int? -> ArrayList() }.add(pair[1])
            graph.computeIfAbsent(pair[1]) { _: Int? -> ArrayList() }.add(pair[0])
        }
        val res = IntArray(graph.size)
        for (entry: Map.Entry<Int, List<Int>> in graph.entries) {
            if (entry.value.size == 1) {
                res[0] = entry.key
                break
            }
        }
        res[1] = graph[res[0]]!![0]
        for (i in 2 until res.size) {
            for (cur: Int in graph[res[i - 1]]!!) {
                if (cur != res[i - 2]) {
                    res[i] = cur
                    break
                }
            }
        }
        return res
    }
}

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