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);
sorted.add(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++) {
graph[tm.get(list[i - 1])].add(tm.get(list[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
seen.add(node)
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]
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