Answer:
Deliberate
Explanation:
Plagiarizing means when some one copy the content of any topic from the sources i.e website, solutions manual, friend manual. And if some one check the solutions and find it that it was copied from somewhere the same we termed as plagiarizing.
In the given situation, if someone copying the content from someone else document so it would be called as deliberate plagiarizing
And if someone did not have any idea about it so we called accidental plagiarizing
The summary of the research - clinical outcomes and survival differences between primary, secondary and concomitants carcinoma in situ of urinary bladder treated with BCG Immunotherapy is given below.
<h3>What is the summary of the above research?</h3>
Urinary bladder carcinoma in situ (CIS) is a flat, high-grade, and aggressive form of urothelial cancer with a high likelihood of development to muscle-invasive illness and metastatic dissemination.
The study's goal was to compare the clinical histories and survival rates of primary, secondary, and concurrent CIS of the bladder.
Learn more about Immunotherapy at brainly.com/question/14493516
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Answer:
Explanation:
The following code is written in Python. It is a recursive function that tests the first and last character of the word and keeps checking to see if each change would create the palindrome. Finally, printing out the minimum number needed to create the palindrome.
import sys
def numOfSwitches(word, start, end):
if (start > end):
return sys.maxsize
if (start == end):
return 0
if (start == end - 1):
if (word[start] == word[end]):
return 0
else:
return 1
if (word[start] == word[end]):
return numOfSwitches(word, start + 1, end - 1)
else:
return (min(numOfSwitches(word, start, end - 1),
numOfSwitches(word, start + 1, end)) + 1)
word = input("Enter a Word: ")
start = 0
end = len(word)-1
print("Number of switches required for palindrome: " + str(numOfSwitches(word, start, end)))
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.
