I think the answer to this would be true
Answer:
- def getCharacterForward(char, key):
- charList = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
-
- if(len(char) > 1):
- return None
- elif(not isinstance(key, int)):
- return -1
- else:
- index = charList.find(char)
- if(index + key <= 25):
- return charList[index + key]
- else:
- return charList[(index + key)% 26]
-
- print(getCharacterForward("C", 4))
- print(getCharacterForward("X", 4))
Explanation:
Firstly, define a charList that includes all uppercase alphabets (Line 2). We presume this program will only handle uppercase characters.
Follow the question requirement and define necessary input validation such as checking if the char is a single character (Line 4). We can do the validation by checking if the length of the char is more than 1, if so, this is not a single character and should return None (Line 5). Next, validate the key by using isinstance function to see if this is an integer. If this is not an integer return -1 (Line 6 - 7).
Otherwise, the program will proceed to find the index of char in the charList using find method (Line 9). Next, we can add the key to index and use the result value to get forwarded character from the charList and return it as output (Line 11).
However, we need to deal a situation that the char is found at close end of the charList and the forward key steps will be out of range of alphabet list. For example the char is X and the key is 4, the four steps forward will result in out of range error. To handle this situation, we can move the last two forward steps from the starting point of the charList. So X move forward 4 will become B. We can implement this logic by having index + key modulus by 26 (Line 13).
We can test the function will passing two sample set of arguments (Line 15 - 16) and we shall get the output as follows:
G
B
In an if...else statement, if the code in the parenthesis of the if statement is true, the code inside its brackets is executed. But if the statement inside the parenthesis is false, all the code within the else statement's brackets is executed instead.
Of course, the example above isn't very useful in this case because true always evaluates to true. Here's another that's a bit more practical:
#include <stdio.h>
int main(void) {
int n = 2;
if(n == 3) { // comparing n with 3 printf("Statement is True!\n");
}
else { // if the first condition is not true, come to this block of code
printf("Statement is False!\n"); } return 0;
}
Output:
Statement is False!
Answer:
The correct answer to the following question will be Option 3 (Professional bureaucracy).
Explanation:
- Professional bureaucracy is evidence that uncentralized organizations can be administrative. Their organizational function is reliable, culminating in "preconceived or repetitive actions, in essence, uniform."It's also very complicated, and so the operators who are doing it should be regulated.
- Mintzberg's organizational framework categorization classifies the information-based organization where services and goods depend as a highly qualified bureaucracy on both the knowledge and expertise of experts.
The other alternatives are not related to the structure of the Mintzberg. So choice 3 is the correct answer.
