Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
A common technique for remembering the order<span> of </span>operations<span> is the abbreviation "PEMDAS", which is turned into the phrase "Please Excuse My Dear Aunt Sally". It stands for "Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction".</span>
Answer:
Cache is the fastest and most expensive, RAM is slower and less expensive, and virtual memory is the slowest and least expensive type.
Explanation:
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
