Problem statement:
Given n processes, each process has a unique PID (process id) and its PPID (parent process id).
Each process only has one parent process, but may have one or more children processes. This is just like a tree structure. Only one process has PPID that is 0, which means this process has no parent process. All the PIDs will be distinct positive integers.
We use two list of integers to represent a list of processes, where the first list contains PID for each process and the second list contains the corresponding PPID.
Now given the two lists, and a PID representing a process you want to kill, return a list of PIDs of processes that will be killed in the end. You should assume that when a process is killed, all its children processes will be killed. No order is required for the final answer.
Example 1:
Input:
pid = [1, 3, 10, 5]
ppid = [3, 0, 5, 3]
kill = 5
Output: [5,10]
Explanation:
3
/ \
1 5
/
10
Kill 5 will also kill 10.
Note:
- The given kill id is guaranteed to be one of the given PIDs.
- n >= 1.
Solution one: TLE
This is the second question of weekly contest 32. It a very good question, tests your basic computer knowledge and programming skills. It comes from real computer world since the process hierarchy is well-known by every computer scientist.
After extracting the information, it is a BFS model. We can naively search the ppid to get its direct children, and no doubt, it is TLE. For each element, we need O(n) time to search its direct children which is at mot the number of n, time complexity will get as high as O(n ^ n), the space complexity is O(1). But, I stupidly solve the problem like this and report a TLE bug.
class Solution {
public:
vector<int> killProcess(vector<int>& pid, vector<int>& ppid, int kill) {
vector<int> proc;
queue<int> que;
que.push(kill);
while(!que.empty()){
int pp_id = que.front();
proc.push_back(pp_id);
que.pop();
for(int i = ; i < ppid.size(); i++){
if(ppid[i] == pp_id){
que.push(pid[i]);
}
}
}
return proc;
}
};
Solution two: BFS + unordered_map(AC)
We need to optimize the solution one. One preprocessing step is to build a hash table indexed by process ID, and get the direct children set by this index. Each time, the time complexity to get the direct children of a process is O(1), the total time complexity is O(n). The space complexity is O(n) as there is a hash table to store the children information of each process.
class Solution {
public:
vector<int> killProcess(vector<int>& pid, vector<int>& ppid, int kill) {
unordered_map<int, vector<int>> hash; // <pid, children_set>
// build hash table which indexed by process ID
for(int i = ; i < ppid.size(); i++){
hash[ppid[i]].push_back(pid[i]);
}
// bfs queue
queue<int> que;
que.push(kill);
vector<int> proc;
// bfs loop
while(!que.empty()){
int pp_id = que.front();
que.pop();
proc.push_back(pp_id);
for(int i = ; i < hash[pp_id].size(); i++){
que.push(hash[pp_id][i]);
}
}
return proc;
}
};
Solution three: DFS + unordered_map(AC)
Same with solution two, this approach employs DFS to search the process.
class Solution {
public:
vector<int> killProcess(vector<int>& pid, vector<int>& ppid, int kill) {
unordered_map<int, vector<int>> hash_table; // <pid, children_set>
// build hash table which indexed by process ID
for(int i = ; i < ppid.size(); i++){
hash_table[ppid[i]].push_back(pid[i]);
}
vector<int> proc;
proc.push_back(kill);
find_process_dfs(proc, hash_table, kill);
return proc;
}
private:
//dfs search
void find_process_dfs(vector<int>& proc, unordered_map<int, vector<int>>& hash_table, int pid){
for(auto process : hash_table[pid]){
proc.push_back(process);
find_process_dfs(proc, hash_table, process);
}
return;
}
};