#636 - Exclusive Time of Functions
Problem Description
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.
Solution
/**
* @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;
};