Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed.
(I'm assuming that := is a typo and you mean just =, by the way)
In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.
This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.
HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES.
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As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2).
(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.
A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
Answer:
maybe blockchain can help them in term of security
Answer:
Explanation:
C++ Code
#include <iostream>
#include <cstdlib>
using namespace std;
int main(){
double hour,minute;
cout<<"Enter Hours :";
cin>>hour;
cout<<"Enter Minutes :";
cin>>minute;
minute = minute+15;
if(minute >=60){
hour++;
minute = minute-60;
}
if(hour>23){
hour = 0;
}
cout<<"Hour: "<< hour<< " Minutes: "<<minute;
return 0;
}
Code Explanation
First take hours and minutes as input. Then add 15 into minutes.
If minutes exceeds from 60 then increment into hours and also remove 60 from minutes as hours already incremented.
Then check if hours are greater then 23 then it means new day is start and it should be 0.
Output
Enter Hours :9
Enter Minutes :46
Hour: 10 Minutes: 1
Answer:
The function is as follows:
def get_win_percentage(self):
return self.team_wins / (self.team_wins + self.team_losses)
Explanation:
This defines the function
def get_win_percentage(self):
This calculates the win percentage and returns it to main
return self.team_wins / (self.team_wins + self.team_losses)
<em>See attachment for complete (and modified) program.</em>
