Last updated on July 18th, 2024 at 03:12 am
Here, We see Exclusive Time of Functions LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.
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Topics
Stack
Companies
Facebook, Uber
Level of Question
Medium
Exclusive Time of Functions LeetCode Solution
Table of Contents
Problem Statement
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.
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 the beginning of time 0, then it executes 2 for units of time and reaches the end of time 1.
Function 1 starts at the beginning of time 2, executes for 4 units of time, and ends at the end of time 5.
Function 0 resumes execution at the beginning of time 6 and executes for 1 unit of time. So function 0 spends 2 + 1 = 3 units of total time executing, and function 1 spends 4 units of total time executing.
Example 2:
Input: n = 1, logs = [“0:start:0″,”0:start:2″,”0:end:5″,”0:start:6″,”0:end:6″,”0:end:7”]
Output: [8]
Explanation:
Function 0 starts at the beginning of time 0, executes for 2 units of time, and recursively calls itself.
Function 0 (recursive call) starts at the beginning of time 2 and executes for 4 units of time.
Function 0 (initial call) resumes execution then immediately calls itself again.
Function 0 (2nd recursive call) starts at the beginning of time 6 and executes for 1 unit of time.
Function 0 (initial call) resumes execution at the beginning of time 7 and executes for 1 unit of time. So function 0 spends 2 + 4 + 1 + 1 = 8 units of total time executing.
Example 3:
Input: n = 2, logs = [“0:start:0″,”0:start:2″,”0:end:5″,”1:start:6″,”1:end:6″,”0:end:7”]
Output: [7,1]
Explanation:
Function 0 starts at the beginning of time 0, executes for 2 units of time, and recursively calls itself.
Function 0 (recursive call) starts at the beginning of time 2 and executes for 4 units of time.
Function 0 (initial call) resumes execution then immediately calls function 1.
Function 1 starts at the beginning of time 6, executes 1 unit of time, and ends at the end of time 6.
Function 0 resumes execution at the beginning of time 6 and executes for 2 units of time. So function 0 spends 2 + 4 + 1 = 7 units of total time executing, and function 1 spends 1 unit of total time executing.
1. Exclusive Time of Functions LeetCode Solution C++
class Solution {
public:
vector<int> exclusiveTime(int n, vector<string>& logs) {
vector<int> ans(n);
stack<pair<int,int>> cur;
int prevTime = 0;
for (auto& s : logs){
int num = stoi(s.substr(0, s.find(':')));
int time = stoi(s.substr(s.rfind(':') + 1));
if (s.find('e') != -1){
ans[num] += time - prevTime + 1;
cur.pop();
prevTime = time + 1;
}
else{
if (!cur.empty()) ans[cur.top().first] += time - prevTime ;
cur.push({num, time});
prevTime = time;
}
}
return ans;
}
};
2. Exclusive Time of Functions LeetCode Solution Java
class Solution {
public int[] exclusiveTime(int n, List<String> logs) {
int[] result = new int[n];
if (n == 0 || logs == null || logs.size() == 0) {
return result;
}
Deque<Integer> stack = new ArrayDeque<>();
int prevTime = 0;
for (String log : logs) {
String[] logParts = log.split(":");
int curTime = Integer.parseInt(logParts[2]);
if ("start".equals(logParts[1])) {
if (!stack.isEmpty()) {
result[stack.peek()] += curTime - prevTime;
}
stack.push(Integer.parseInt(logParts[0]));
prevTime = curTime;
} else {
result[stack.pop()] += curTime - prevTime + 1;
prevTime = curTime + 1;
}
}
return result;
}
}
3. Exclusive Time of Functions LeetCode Solution JavaScript
var exclusiveTime = function(n, logs) {
const sums = new Array(n).fill(0);
const stack = [];
let prevTime;
logs.forEach(log => {
const details = log.split(':');
const id = parseInt(details[0]);
const point = details[1];
const time = parseInt(details[2]);
if (point === 'start') {
if (stack.length > 0) {
let prevFn = stack[stack.length - 1];
sums[prevFn] += (time - prevTime);
}
stack.push(id);
prevTime = time;
} else {
const last = stack.pop();
sums[last] += (time - prevTime + 1);
prevTime = time + 1;
}
});
return sums;
};
4. Exclusive Time of Functions LeetCode Solution Python
class Solution(object):
def exclusiveTime(self, n, logs):
helper = lambda log: (int(log[0]), log[1], int(log[2]))
logs = [helper(log.split(':')) for log in logs]
ans, s = [0] * n, []
for (i, status, timestamp) in logs:
if status == 'start':
if s: ans[s[-1][0]] += timestamp - s[-1][1]
s.append([i, timestamp])
else:
ans[i] += timestamp - s.pop()[1] + 1
if s: s[-1][1] = timestamp+1
return ans