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Ne4ueva [31]
3 years ago
12

Which of the following scenarios might indicate that you have been a victim of identity theft?A)Your credit report displays acco

unts you did not open.B)You do not pay your credit card bill on timeC)Both A and BD)neither A or B
Computers and Technology
1 answer:
victus00 [196]3 years ago
7 0

Answer:

Option A is the correct answer for the above question.

Explanation:

A Victim is a person, who faces the problem of a criminal person. Here in the question, the scenario is that a user is a victim of the identity of being theft by some other person. So the user can get known by the help of option a, which suggests that the user's credit card report display something which is not done by him then it can be the scenario, where the user can understand that he is a victim. hence option a is the correct answer where the other option is not because--

  • Option B suggests the user does not pay the bill on time but it is a general case because the bill is paid by the user is in every month not a single time.
  • Option C suggests both options (A and B) are correct, but option b is not the correct.
  • Option D suggests that no option is valid from the above but option A is the correct.
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Suppose you have two arrays of ints, arr1 and arr2, each containing ints that are sorted in ascending order. Write a static meth
telo118 [61]
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++;
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}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
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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.


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WHY IS BRAINLY NOTIFICATIONS LIKE THIS?????
nikdorinn [45]

Answer:

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Explanation:

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