# Recover Original List From Subsequences - Google Top Interview Questions

### Problem Statement :

```Given a two-dimensional list of integers lists, where each list is a subsequence of an original list, return the original list.

The original list contains unique numbers. There is guaranteed to be one unique solution.

Constraints

n * m ≤ 100,000 where n and m are the number of rows and columns in lists

Example 1

Input

lists = [

[1, 3],

[2, 3],

[1, 2]

]

Output

[1, 2, 3]

Example 2

Input

lists = [

[1, 2, 4],

[2, 3, 4]

]

Output

[1, 2, 3, 4]

Example 3

Input

lists = [

[1, 2, 3]

]

Output

[1, 2, 3]```

### Solution :

```                        ```Solution in C++ :

vector<int> solve(vector<vector<int>>& lists) {
unordered_map<int, vector<int>> edges;
unordered_set<int> vertices;
unordered_map<int, int> numParents;
for (auto& list : lists) {
vertices.insert(list[0]);
for (int i = 1; i < list.size(); i++) {
vertices.insert(list[i]);
edges[list[i - 1]].push_back(list[i]);
numParents[list[i]]++;
}
}
vector<int> ret;
for (int vertex : vertices) {
if (numParents[vertex] == 0) {
ret.push_back(vertex);
}
}
for (int i = 0; i < ret.size(); i++) {
for (int child : edges[ret[i]]) {
if (--numParents[child] == 0) {
ret.push_back(child);
}
}
}
return ret;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int[] solve(int[][] lists) {
// normalize the numbers
TreeSet<Integer> nums = new TreeSet<Integer>();
for (int[] list : lists) {
for (int l : list) nums.add(l);
}
int N = 0;
TreeMap<Integer, Integer> tm = new TreeMap<Integer, Integer>();
ArrayList<Integer> sorted = new ArrayList<Integer>();
for (int n : nums) {
tm.put(n, N);
N++;
}

ArrayList<Integer>[] graph = new ArrayList[N];
int[] prev = new int[N];
for (int i = 0; i < N; i++) graph[i] = new ArrayList<Integer>();
for (int[] list : lists) {
for (int i = 1; i < list.length; i++) {
prev[tm.get(list[i])] += 1;
}
}

int cur = -1;
for (int i = 0; i < N; i++) {
if (prev[i] == 0) {
cur = i;
break;
}
}

int[] ans = new int[N];
int index = 0;
while (cur >= 0) {
ans[index] = sorted.get(cur);
index++;
for (int n : graph[cur]) {
prev[n] -= 1;
}
int next = -1;
for (int n : graph[cur]) {
if (prev[n] == 0) {
next = n;
break;
}
}
cur = next;
}
return ans;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, A):
graph = defaultdict(list)
for row in A:
for i in range(len(row) - 1):
graph[row[i]].append(row[i + 1])
graph[row[i + 1]]

seen = set()

def dfs(node):
if node in seen:
return
for nei in graph[node]:
if nei not in seen:
dfs(nei)
post.append(node)

post = []
for u in graph:
dfs(u)
return post[::-1]```
```

## Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink

## Insert a Node at the head of a Linked List

Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below

## Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

## Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

## Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio