0636-exclusive-time-of-functions.js

/**
 * 636. Exclusive Time of Functions
 * https://leetcode.com/problems/exclusive-time-of-functions/
 * Difficulty: Medium
 *
 * On a single-threaded CPU, we execute a program containing n functions. Each function has a
 * unique ID between 0 and n-1.
 *
 * Function calls are stored in a call stack: when a function call starts, its ID is pushed
 * onto the stack, and when a function call ends, its ID is popped off the stack. The function
 * whose ID is at the top of the stack is the current function being executed. Each time a
 * function starts or ends, we write a log with the ID, whether it started or ended, and the
 * timestamp.
 *
 * You are given a list logs, where logs[i] represents the ith log message formatted as a string
 * "{function_id}:{"start" | "end"}:{timestamp}". For example, "0:start:3" means a function call
 * with function ID 0 started at the beginning of timestamp 3, and "1:end:2" means a function
 * call with function ID 1 ended at the end of timestamp 2. Note that a function can be called
 * multiple times, possibly recursively.
 *
 * A function's exclusive time is the sum of execution times for all function calls in the program.
 * For example, if a function is called twice, one call executing for 2 time units and another
 * call executing for 1 time unit, the exclusive time is 2 + 1 = 3.
 *
 * Return the exclusive time of each function in an array, where the value at the ith index
 * represents the exclusive time for the function with ID i.
 */

/**
 * @param {number} n
 * @param {string[]} logs
 * @return {number[]}
 */
var exclusiveTime = function(n, logs) {
  const stack = [];
  const result = new Array(n).fill(0);
  let previousTime = 0;

  for (const log of logs) {
    const [id, action, time] = log.split(':');
    const currentTime = +time;

    if (action === 'start') {
      if (stack.length) {
        result[stack[stack.length - 1]] += currentTime - previousTime;
      }
      stack.push(+id);
      previousTime = currentTime;
    } else {
      result[stack.pop()] += currentTime - previousTime + 1;
      previousTime = currentTime + 1;
    }
  }

  return result;
};