Answer: Yes
Explanation:
It would force the company to either enforce the policy and fire all of its employees.
Answer:
To obtain substantial performance enhancements, we must exploit ILP across multiple basic blocks.
Explanation:
Answer:
- def check_subset(l1, l2):
- status = False
- count = 0
- if(len(l1) > len(l2)):
- for x in l2:
- for y in l1:
- if x == y:
- count += 1
-
- if(count == len(l2)):
- return True
- else:
- return False
-
- else:
- for x in l1:
- for y in l2:
- if x==y:
- count += 1
-
- if(count == len(l1)):
- return True
- else:
- return False
-
- print(check_subset([1,4,6], [1,2,3,4,5,6]))
- print(check_subset([2,5,7,9,8], [7,8]))
- print(check_subset([1, 5, 7], [1,4,6,78,12]))
Explanation:
The key idea of this solution is to create a count variable to track the number of the elements in a shorter list whose value can be found in another longer list.
Firstly, we need to check which list is shorter (Line 4). If the list 2 is shorter, we need to traverse through the list 2 in an outer loop (Line 5) and then create another inner loop to traverse through the longer list 1 (Line 6). If the current x value from list 2 is matched any value in list 1, increment the count variable by 1. After finishing the outer loop and inner loop, we shall be able to get the total count of elements in list 2 which can also be found in list 1. If the count is equal to the length of list 2, it means all elements in the list 2 are found in the list 1 and therefore it is a subset of list 1 and return true (Line 10-11) otherwise return false.
The similar process is applied to the situation where the list 1 is shorter than list 2 (Line 15-24)
If we test our function using three pairs of input lists (Line 26-28), we shall get the output as follows:
True
True
False
Answer: Letter recognition
Fine motor skills
Isolated use of each finger
Bilateral coordination (using two hands together)
Eye-hand coordination
Explanation:
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
