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Exclusive Time of Functions in C++
Suppose on a single threaded CPU, we execute some functions. Now each function has a unique id between 0 and N-1. We will store logs in timestamp order that describe when a function is entered or exited.
Here each log is a string written this format: "{function_id}:{"start" | "end"}:{timestamp}". For example, if the string is like "0:start:3" this means that the function with id 0 started at the beginning of timestamp 3. "1:end:2" means the function with id 1 ended at the end of timestamp 2. A function's exclusive time is the number of units of time spent in this function.
So if the input is like n = 2 and logs = ["0:start:0","1:start:2","1:end:5","0:end:6"], then the output will be [3,4]. This is because function 0 starts at the beginning of time 0, then it executes 2 units of time and reaches the end of time 1. After that function 1 starts at the beginning of time 2, executes 4 units of time and ends at time 5. The function 0 is running again at the beginning of time 6, and also ends at the end of time 6, thus executing for 1 unit of time. So we can see that the function 0 spends 2 + 1 = 3 units of total time executing, and function 1 spends 4 units of total time executing.
To solve this, we will follow these steps −
Define an array ret of size n, define stack st
j := 0, prev := 0
for i in range 0 to size of log array – 1
temp := logs[i], j := 0, id := 0, num := 0, type := empty string
while temp[j] is not a colon
id := id * 10 + temp[j] as number
increase j by 1
increase j by 1
while temp[j] is not a colon
type := type concatenate temp[j]
increase j by 1
increase j by 1
while j < size of temp
num := num * 10 + temp[j] as number
increase j by 1
if type = start, then
if st is not empty
increase ret[stack top element] by num – prev
insert d into st, prev := num
otherwise
x := top of st, and delete top of stack
ret[x] := ret[x] + (num + 1) – prev
prev := num + 1
return ret
Example(C++)
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
class Solution {
public:
vector<int> exclusiveTime(int n, vector<string>& logs) {
vector <int> ret(n);
stack <int> st;
int id, num;
int j = 0;
string temp;
string type;
int prev = 0;
for(int i = 0; i < logs.size(); i++){
temp = logs[i];
j = 0;
id = 0;
num = 0;
type = "";
while(temp[j] != ':'){
id = id * 10 + (temp[j] - '0');
j++;
}
j++;
while(temp[j] != ':'){
type += temp[j];
j++;
}
j++;
while(j < temp.size()){
num = num * 10 + temp[j] - '0';
j++;
}
if(type == "start"){
if(!st.empty()){
ret[st.top()] += num - prev;
}
st.push(id);
prev = num;
} else {
int x = st.top();
st.pop();
ret[x] += (num + 1) - prev;
prev = num + 1;
}
}
return ret;
}
};
main(){
vector<string> v = {"0:start:0","1:start:2","1:end:5","0:end:6"};
Solution ob;
print_vector(ob.exclusiveTime(2, v));
}Input
2 ["0:start:0","1:start:2","1:end:5","0:end:6"]
Output
[3, 4, ]
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