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.
<span>Computer design replaced (B) models draw or created by hand. Technology nowadays has a big contribution in terms of planning and designing a building or a structure without burning your eyebrow facing that paper and handling your pen. Computer designs are used to make the design more accurate and more precise compared to traditional hand drawn designs.</span>
Answer:
"Mesh topology" is the correct answer.
Explanation:
- A mesh topology seems to be a network configuration where there has been an interconnection between each hardware and communications system. This topology configuration allows the distribution of many of these signals, although one of the connexons starts going down.
- This topology needs to connect each destination to any other end destination, thus creating a completely pointless channel.
The answer is b. Susan has an appointment with he co-worker on the first Monday of every month. This scenario would benefit from being set up as a recurring appointment, since it requires the same thing to be done every month on the same day. Susan is likely to forget this appointment every month, so a reminder like this would be useful for her. For the other cases, a recurring appointment would not be of much benefit since they do not follow a particular pattern time-wise.
To get the number ot decoder output:
2^n; where n is number of input
2^4
2 × 2 × 2 × 2 = 16 outputs
