Answer:
d. "Columbus, Wisconsin"
Explanation:
mark me brainliest please
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.
Answer:
True is the correct answer for the above question.
Explanation:
- When any document file is opened then every point has some particular address. so there is a pointer which states that where the read operation and the write operation is going on.
- When any person writes any program to read a file or write a file then there is a need for some variable that is pointed for the reading and the write operation.
- The document which is used for the write data or read data is also designed and maintained by some software.
- Hence we can say that there are needs of some variable that point the operation of the file and it is also stated from the above question. Hence the above question statement is a true statement.
<span>C. 3
Due to the different speeds of P and S waves, a single seismometers can determine the distance to an earthquake. So, for a single station, the localization is any point on a circle around the station. With 2 stations, you'll have two circles that intersect at two points. The 3rd station is needed in order to determine which of the 2 points is the actual earthquake.</span>
Yes , it’s true. In a known-plaintext attack (kpa), the cryptanalyst can only view a small portion of encrypted data, and he or she has no control over what that data might be.
The attacker also has access to one or more pairs of plaintext/ciphertext in a Known Plaintext Attack (KPA). Specifically, consider the scenario where key and plaintext were used to derive the ciphertext (either of which the attacker is trying to find). The attacker is also aware of what are the locations of the output from key encrypting. That is, the assailant is aware of a pair. They might be familiar with further pairings (obtained with the same key).
A straightforward illustration would be if the unencrypted messages had a set expiration date after which they would become publicly available. such as the location of a planned public event. The coordinates are encrypted and kept secret prior to the event. But when the incident occurs, the attacker has discovered the value of the coordinates /plaintext while the coordinates were decrypted (without knowing the key).
In general, a cipher is easier to break the more plaintext/ciphertext pairs that are known.
To learn more about Plaintext Attack click here:
brainly.com/question/28445346
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