Answer:
d) either e1 or e5
Explanation:
Here, the instruction i1 goes ahead in trying to open the given file through an input stream buffer reader. If the given file name is wrong, it will indicate that an e1 file is not found or if any other IO errors due to invalid stream, no disc in drive e5 IO exception will be drawn.
Answer:
What Sherman needs to configure is:
RIPv2 class D of 240.
Explanation:
Multicast messages are usually dispatched to a select group of hosts on a network and require acknowledgement of the recipients. RIPv2 is a router-based internet protocol for exchanging routing information to the IP address 224.0. 0.9 on a network. It determine the most efficient way to route data on a network and to prevent routing loops.
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.
Explanation:
1. How many computers do you want to connect or how big the network should be?
This would tell us what kind of a network need to be built. It can be LAN/MAN/WAN
2. The location where network needs to be built?
We have to check the geographic condition too before creating a network
3. What is the budget?
Based on the budget only, we can decide the wires to be used if require we can negotiate the budget so that we can create effective network
4. Will I get an additional resources to work?
This is essential to estimate the time that is required to complete the task.
5. When the project needs to be completed?
This is crucial because a business might be planned thereafter.
