https://leetcode.com/problems/exclusive-time-of-functions
Use a stack to mimic a function call stack. Stack top will always yield the currently executing fn.
Also keep track of the currently exec function's start/restart timestamp.
Time complexity : . We increment the time till all the functions are done with the execution. Here, refers to the end time of the last function in the list.
Given the running logs of n functions that are executed in a nonpreemptive single threaded CPU, find the exclusive time of these functions.
Each function has a unique id, start from 0 to n-1. A function may be called recursively or by another function.
A log is a string has this format :
function_id:start_or_end:timestamp. For example, "0:start:0" means function 0 starts from the very beginning of time 0. "0:end:0" means function 0 ends to the very end of time 0.
Exclusive time of a function is defined as the time spent within this function, the time spent by calling other functions should not be considered as this function's exclusive time. You should return the exclusive time of each function sorted by their function id.
Example 1:
Input: n = 2 logs = ["0:start:0", "1:start:2", "1:end:5", "0:end:6"] Output:[3, 4] Explanation: Function 0 starts at time 0, then it executes 2 units of time and reaches the end of time 1. Now function 0 calls function 1, function 1 starts at time 2, executes 4 units of time and end at time 5. Function 0 is running again at time 6, and also end at the time 6, thus executes 1 unit of time. So function 0 totally execute 2 + 1 = 3 units of time, and function 1 totally execute 4 units of time.
Note:
- Input logs will be sorted by timestamp, NOT log id.
- Your output should be sorted by function id, which means the 0th element of your output corresponds to the exclusive time of function 0.
- Two functions won't start or end at the same time.
- Functions could be called recursively, and will always end.
- 1 <= n <= 100
public int[] exclusiveTime(int n, List<String> logs) {
int[] res = new int[n];
Stack<Integer> stack = new Stack<>();
int prevTime = 0;
for (String log : logs) {
String[] parts = log.split(":");
if (!stack.isEmpty()) res[stack.peek()] += Integer.parseInt(parts[2]) - prevTime;
prevTime = Integer.parseInt(parts[2]);
if (parts[1].equals("start")) stack.push(Integer.parseInt(parts[0]));
else {
res[stack.pop()]++;
prevTime++;
}
}
return res;
}public int[] exclusiveTime(int n, List<String> logs) {
// separate time to several intervals, add interval to their function
int[] result = new int[n];
Stack<Integer> st = new Stack<>();
int pre = 0;
// pre means the start of the interval
for(String log: logs) {
String[] arr = log.split(":");
if(arr[1].equals("start")) {
if(!st.isEmpty()) result[st.peek()] += Integer.parseInt(arr[2]) - pre;
// arr[2] is the start of next interval, doesn't belong to current interval.
st.push(Integer.parseInt(arr[0]));
pre = Integer.parseInt(arr[2]);
} else {
result[st.pop()] += Integer.parseInt(arr[2]) - pre + 1;
// arr[2] is end of current interval, belong to current interval. That's why we have +1 here
pre = Integer.parseInt(arr[2]) + 1;
// pre means the start of next interval, so we need to +1
}
}
return result;
}
We examine two approaches - both will be stack based.
In a more conventional approach, let's look between adjacent events, with duration
time - prev_time. If we started a function, and we have a function in the background, then it was running during this time. Otherwise, we ended the function that is most recent in our stack.public int[] exclusiveTime(int n, List<String> logs) {
// separate time to several intervals, add interval to their function
int[] result = new int[n];
Stack<Integer> st = new Stack<>();
int pre = 0;
// pre means the start of the interval
for(String log: logs) {
String[] arr = log.split(":");
if(arr[1].equals("start")) {
if(!st.isEmpty()) result[st.peek()] += Integer.parseInt(arr[2]) - pre;
// arr[2] is the start of next interval, doesn't belong to current interval.
st.push(Integer.parseInt(arr[0]));
pre = Integer.parseInt(arr[2]);
} else {
result[st.pop()] += Integer.parseInt(arr[2]) - pre + 1;
// arr[2] is end of current interval, belong to current interval. That's why we have +1 here
pre = Integer.parseInt(arr[2]) + 1;
// pre means the start of next interval, so we need to +1
}
}
return result;
}Use a stack to mimic a function call stack. Stack top will always yield the currently executing fn.
Also keep track of the currently exec function's start/restart timestamp.
public int[] exclusiveTime(int n, List<String> logs) {
int[] result = new int[n];
Stack<Integer> stack = new Stack<>();
String[] log0 = logs.get(0).split(":");
stack.push(Integer.parseInt(log0[0]));
int currExStartTS = Integer.parseInt(log0[2]);
for(int i = 1; i < logs.size(); i++) {
String[] logi = logs.get(i).split(":");
int logi_fn = Integer.parseInt(logi[0]);
int logi_ts = Integer.parseInt(logi[2]);
if(logi[1].equals("start")) {
if(!stack.isEmpty()) { //check if there is no currently executing fn
int currFn = stack.peek();
result[currFn] += logi_ts - currExStartTS;
}
currExStartTS = logi_ts;
stack.push(logi_fn);
}
else {
int endedFn = stack.pop();
result[endedFn] += logi_ts - currExStartTS + 1;
currExStartTS = logi_ts + 1;
}
}
return result;
}Time complexity : . We increment the time till all the functions are done with the execution. Here, refers to the end time of the last function in the list.
public int[] exclusiveTime(int n, List < String > logs) { Stack < Integer > stack = new Stack < > (); int[] res = new int[n]; String[] s = logs.get(0).split(":"); stack.push(Integer.parseInt(s[0])); int i = 1, time = Integer.parseInt(s[2]); while (i < logs.size()) { s = logs.get(i).split(":"); while (time < Integer.parseInt(s[2])) { res[stack.peek()]++; time++; } if (s[1].equals("start")) stack.push(Integer.parseInt(s[0])); else { res[stack.peek()]++; time++; stack.pop(); } i++; } return res; }https://mikecoder.github.io/oj-code/2017/07/16/ExclusiveTimeofFunctions/
class Task {
int id;
int startTime;
int endTime;
int totalTime = -1;
ArrayList<Task> nextTasks = new ArrayList<>();
Task parentTask = null;
Task(int id, int startTime) {
this.id = id;
this.startTime = startTime;
}
}
public int[] exclusiveTime(int n, List<String> logs) {
int res[] = new int[n];
Task rootTask = new Task(-1, -1);
calcTask(rootTask, null, 0, logs);
countTime(rootTask);
addRes(rootTask, res);
return res;
}
private void addRes(Task task, int res[]) {
if (task.id != -1) {
res[task.id] += task.totalTime;
}
for (int i = 0; i < task.nextTasks.size(); i++) {
addRes(task.nextTasks.get(i), res);
}
}
private void countTime(Task currentTask) {
currentTask.totalTime = currentTask.endTime - currentTask.startTime + 1;
if (currentTask.nextTasks.size() != 0) {
for (int i = 0; i < currentTask.nextTasks.size(); i++) {
currentTask.totalTime = currentTask.totalTime - (currentTask.nextTasks.get(i).endTime - currentTask.nextTasks.get(i).startTime + 1);
countTime(currentTask.nextTasks.get(i));
}
}
}
private void calcTask(Task currentTask, Task parentTask, int idx, List<String> logs) {
if (idx == logs.size()) {
return;
}
String log = logs.get(idx);
int logInfo[] = parse(log);
if (log.contains("start")) {
Task newTask = new Task(logInfo[0], logInfo[1]);
newTask.parentTask = currentTask;
currentTask.nextTasks.add(newTask);
calcTask(newTask, currentTask, idx + 1, logs);
} else {
currentTask.endTime = logInfo[1];
calcTask(parentTask, parentTask.parentTask, idx + 1, logs);
}
}
private int[] parse(String log) {
int logId = Integer.valueOf(log.substring(0, log.indexOf(":")));
int timeStamp;
if (log.contains("start")) {
timeStamp = Integer.valueOf(log.substring(log.indexOf("t:") + 2));
} else {
timeStamp = Integer.valueOf(log.substring(log.indexOf("d:") + 2));
}
int res[] = new int[2];
res[0] = logId;
res[1] = timeStamp;
return res;
}
