Answer:
def printInfo(some_dict):
print(len(some_dict['locations']), "LOCATIONS")
for location in some_dict['locations']:
print(location)
print()
print(len(some_dict['instructors']), "INSTRUCTORS")
for location in some_dict['instructors']:
print(location)
seattle = {
'locations': ['San Jose', 'Seattle', 'Dallas', 'Chicago', 'Tulsa', 'DC', 'Burbank'],
'instructors': ['Michael', 'Amy', 'Eduardo', 'Josh', 'Graham', 'Patrick', 'Minh', 'Devon']
}
printInfo(seattle)
Explanation:
Answer:
O(N!), O(2N), O(N2), O(N), O(logN)
Explanation:
N! grows faster than any exponential functions, leave alone polynomials and logarithm. so O( N! ) would be slowest.
2^N would be bigger than N². Any exponential functions are slower than polynomial. So O( 2^N ) is next slowest.
Rest of them should be easier.
N² is slower than N and N is slower than logN as you can check in a graphing calculator.
NOTE: It is just nitpick but big-Oh is not necessary about speed / running time ( many programmers treat it like that anyway ) but rather how the time taken for an algorithm increase as the size of the input increases. Subtle difference.
The most efficient thing to do would be to provide training as firing or doing nothing halts productivity indefinitely while training eventually will finish
Solution :
#include 
#include 
#include 
//Converts 
to binary string.
* hexadecimal
Binary(char* hexdec)
{
long 
= 0;
char *string = 
(sizeof(char) * 9);
while (hexdec[i]) {
//Simply assign binary string for each hex char.
switch (hexdec[i]) {
strcat(string, "0000");
break;
strcat(string, "0001");
break;
strcat(string, "0010");
break;
strcat(string, "0011");
break;
strcat(string, "0100");
break;
strcat(string, "0101");
break;
strcat(string, "0110");
break;
strcat(string, "0111");
break;
strcat(string, "1000");
break;
strcat(string, "1001");
break;
case 'A':
case 'a':
strcat(string, "1010");
break;
case 'B':
case 'b':
strcat(string, "1011");
break;
case 'C':
case 'c':
strcat(string, "1100");
break;
case 'D':
case 'd':
strcat(string, "1101");
break;
case 'E':
case 'e':
strcat(string, "1110");
break;
case 'F':
case 'f':
strcat(string, "1111");
break;
default:
printf("\nInvalid hexadecimal digit %c",
hexdec[i]);
string="-1" ;
}
i++;
}
return string;
}
int main()
{ //Take 2 strings
char *str1 =hexadecimalToBinary("FA") ;
char *str2 =hexadecimalToBinary("12") ;
//Input 2 numbers p and n.
int p,n;
scanf("%d",&p);
scanf("%d",&n);
//keep j as length of str2
int j=strlen(str2),i;
//Now replace n digits after p of str1
for(i=0;i<n;i++){
str1[p+i]=str2[j-1-i];
}
//Now, i have used c library strtol
long ans = strtol(str1, NULL, 2);
//print result.
printf("%lx",ans);
return 0;
}
Answer:
Option (C) is the correct option to the following question.
Explanation:
The following option is correct because the unit testing is the process of testing a single unit of software at a time, which means the testing of each and every program separately.
In simple words, Unit testing a process of testing in which the developer executes the single method or a function, statements or loop in the program of the software to checking is it working fine or not.
