582. Kill Process

Posted 2020-09-18 蓝色地中海

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:

1. The given kill id is guaranteed to be one of the given PIDs.
2. n >= 1.

Solution one:

This is the second question of weekly contest 32. It a very good question, and can test your basic computer knowledgement and programming skills. It comes from real computer world since the process hierarchy is well-known by every computer scientist. 

This problem abstract the problem and form a BFS model. We can naively search the ppid to get it`s direct children, and no doubt, it is TLE, obviously it is not a good time complexity. For each element, we need O(n) time to search its direct children. But, I stupidly solve the problem like this and report an 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 = 0; i < ppid.size(); i++){
                if(ppid[i] == pp_id){
                    que.push(pid[i]);
                }
            }  
        }
        return proc;
    }
};

Solution two:

We need to optimize the solution one. One extra 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). This greatly reduced the time complexity.

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 = 0; 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 = 0; i < hash[pp_id].size(); i++){
                que.push(hash[pp_id][i]);
            }
        }
        return proc;
    }
};