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| 1 | +package com.fishercoder.solutions; |
| 2 | + |
| 3 | +import java.util.Deque; |
| 4 | +import java.util.LinkedList; |
| 5 | +import java.util.List; |
| 6 | + |
| 7 | +/** |
| 8 | + * 636. Exclusive Time of Functions |
| 9 | + * |
| 10 | + * Given the running logs of n functions that are executed in a nonpreemptive single threaded CPU, |
| 11 | + * find the exclusive time of these functions. |
| 12 | + * Each function has a unique id, start from 0 to n-1. |
| 13 | + * A function may be called recursively or by another function. |
| 14 | + * A log is a string has this format : function_id:start_or_end:timestamp. |
| 15 | + * For example, "0:start:0" means function 0 starts from the very beginning of time 0. |
| 16 | + * "0:end:0" means function 0 ends to the very end of time 0. |
| 17 | + * Exclusive time of a function is defined as the time spent within this function, |
| 18 | + * the time spent by calling other functions should not be considered as this function's exclusive time. |
| 19 | + * You should return the exclusive time of each function sorted by their function id. |
| 20 | +
|
| 21 | + Example 1: |
| 22 | + Input: |
| 23 | + n = 2 |
| 24 | + logs = |
| 25 | + ["0:start:0", |
| 26 | + "1:start:2", |
| 27 | + "1:end:5", |
| 28 | + "0:end:6"] |
| 29 | + Output:[3, 4] |
| 30 | + Explanation: |
| 31 | + Function 0 starts at time 0, then it executes 2 units of time and reaches the end of time 1. |
| 32 | + Now function 0 calls function 1, function 1 starts at time 2, executes 4 units of time and end at time 5. |
| 33 | + Function 0 is running again at time 6, and also end at the time 6, thus executes 1 unit of time. |
| 34 | + So function 0 totally execute 2 + 1 = 3 units of time, and function 1 totally execute 4 units of time. |
| 35 | +
|
| 36 | + Note: |
| 37 | + Input logs will be sorted by timestamp, NOT log id. |
| 38 | + Your output should be sorted by function id, which means the 0th element of your output corresponds to the exclusive time of function 0. |
| 39 | + Two functions won't start or end at the same time. |
| 40 | + Functions could be called recursively, and will always end. |
| 41 | + 1 <= n <= 100 |
| 42 | + */ |
| 43 | +public class _636 { |
| 44 | + /**Based on the example, it's difficult to see how function 2 executes 4 units of time, actually |
| 45 | + * we can add 1 to all end times to make it easier to understand and AC'ed.*/ |
| 46 | + public int[] exclusiveTime(int n, List<String> logs) { |
| 47 | + /**Stack is the way to go: |
| 48 | + * we keep pushing the logId onto the stack whenever we just encounter this logId's start timestamp, |
| 49 | + * we'll pop this logId only when we encounter this logId's end timestamp. |
| 50 | + * Meanwhile, we keep a counter called prevTime, |
| 51 | + * whenever the stack is not empty, we'll always deduct prevTime from the last logId on the stack.*/ |
| 52 | + Deque<Integer> stack = new LinkedList<>(); |
| 53 | + int[] result = new int[n]; |
| 54 | + int prevTime = 0; |
| 55 | + for (String log : logs) { |
| 56 | + String[] parts = log.split(":"); |
| 57 | + if (!stack.isEmpty()) { |
| 58 | + result[stack.peek()] += Integer.parseInt(parts[2]) - prevTime; |
| 59 | + } |
| 60 | + prevTime = Integer.parseInt(parts[2]); |
| 61 | + if (parts[1].equals("start")) { |
| 62 | + stack.addFirst(Integer.parseInt(parts[0]));//i.e. stack.push() |
| 63 | + } else { |
| 64 | + prevTime++; |
| 65 | + //remember to have result pluse 1 to match the problem AC criteria |
| 66 | + result[stack.pollFirst()]++;//i.e. stack.pop() |
| 67 | + } |
| 68 | + } |
| 69 | + return result; |
| 70 | + } |
| 71 | +} |
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