Answer:
Answered below
Explanation:
Imperative programming paradigm is a paradigm in which the programmer tells the machine how to change its state. It is divided into procedural programming, where programs are grouped as procedures, and object-oriented, where programs are grouped together with the properties they operate on.
Imperative paradigm is efficient, familiar and popular compared to others.
Functional programming paradigm implements all instructions as functions, in a mathematical sense. They take on an argument and return a single solution. The advantage of this, compared to other paradigms, is its abstraction and independence.
Logical programming, like functional programming, incorporates a declarative approach to problem solving. All known facts are first established before queries are made. Advantage of this paradigm is that programming steps are kept to a minimum.
The correct answer would be Forums
Answer:
35
Explanation:
We will be going inside B array, because he was in second place in array E,and will be the first element of array B, so it's 35
E can be understanded as:
E=[[21, 'dog', 'red'],[35, 'cat', 'blue'],[12, 'fish', 'green']], so, you can see array E as array of arrays or so-called two-dimensional array
Complete Question:
1. A wireless technology standard for exchanging data over short distances
2. A particular brand of mobile phone/PDA
3. A network that operates over a limited distance, usually for one or a few users
1. Bluetooth
2. PAN
3. Blackberry
Answer:
1. Bluetooth 2. Blackberry . 3. PAN
Explanation:
1. Bluetooth is a wireless technology standard, used in order to exchange data between mobile devices, like smartphones, tablets, headsets, wearables, over short distances in a one-to-one fashion (which means that it is not possible to build a network based in Bluetooth).
2. Blackberry is a brand of mobile phones/PDAs, very popular a decade ago, because it was the first one to allow mobile users to access e-mails and messages from anywhere, at any time.
3. PAN (Personal Area Network) is an ad-hoc network that it is only available for data exchange at a very short distance, within the reach of a person, i.e. a few meters as a maximum.
It is thought to allow someone to interact with his nearest environment (laptop, tablet, PDA) and it can be wireless (like Bluetooth) or wired (via USB cables).
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.
