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Problem of the day

This time we are executing a program containing ‘N’ functions on a single-threaded CPU. Each function has a unique ‘ID’ between 0 and (N-1) and each time it starts or ends, we write a log with the ID, whether it started or ended, and the TIMESTAMP.

You are given a 2D array of integers containing information about ‘L’ logs where ith column represents the ith log message. Each column contains 3 rows to describe the ith log where,

```
1st Row - represents the ID of the function.
2nd Row - represents whether the function has started or ended where 1 denotes the start and -1 denotes the end.
3rd Row - represents the TIMESTAMP of the log.
```

You are required to return an array where the value at the ith index represents the exclusive time for the function with ID ‘i’.

Note:

```
1. The exclusive time of a function is the sum of execution times for all calls of that function.
2. A function can be called multiple times, possibly recursively.
3. No two events will happen at the same time where an event denotes either a start or end of a function call. This basically means no two logs have the same timestamp.
4. Each function has an end log for each start log.
```

For Example:

```
Consider the following input
0 1 1 1 2 2 1 0
1 1 1 -1 1 -1 -1 -1
0 2 5 7 8 10 11 14
```

```
Thus, we return [5, 7, 3] as a process with ID 0 has taken 5 units of time and process with ID 1 has taken 7 units of time and process ID 2 has taken 3 units of time. A process’s exclusive time is the sum of exclusive times for all function calls in the program. As process Id 1 has called itself so exclusive time is the sum of exclusive times(5 + 2).
```

Detailed explanation

```
1 <= T <= 10
1 <= N <= 10 ^ 2
2 <= L <= 5 * (10 ^ 2) and L is even.
0 <= TIMESTAMP <= 10 ^ 9
Time Limit: 1 sec.
```

```
1
2 4
0 1 1 0
1 1 -1 -1
0 2 5 6
```

```
3 4
```

```
Function 0 starts at the beginning of timestamp 0, then it executes for 2 units of time and reaches the end of timestamp 1.
Function 1 starts at the beginning of timestamp 2, executes for 4 units of time, and ends at the end of timestamp 5.
Function 0 resumes execution at the beginning of timestamp 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.
```

```
1
1 6
0 0 0 0 0 0
1 1 -1 1 -1 -1
0 2 5 6 6 7
```

```
8
```