Answer:
Explanation:
( n cards are there initially )
we pick out the first card in random it takes n-1 comparisons to figure out
its Equivalence card - n-1 steps
Two cards have been eliminated ( this leaves us with 2 and n-2 cards)
we pick out the 2nd card in random it takes n-3 comparisons to figure out
its Equivalence card - n-3 steps
we continue to do this.. till all cards are exhausted ( leaves us with 2
and n-4 cards again)
the last comparison will
have
- n-(n-3)
the sum of all these steps - (n-1) + (n-3) + (n-5) + .........+
(n-(n-3))
if you draw this in the form of a tree.
n - n
2
n-2 - n
2
n-4 - n-2
2
n-6 - n-4
2
n-8 - n- 6
the height of the tree will be log n , sum @ each level is at most n
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.
Answer:
The answer is Magnetic
Explanation:
Magnetic medium is a secondary storage medium that uses magnetic techniques to store and retrieve data on disks or tapes coated with magnetically sensitive materials.
Magnetic based drives, for example:- Hard disk. Floppy Disk. Magnetic tape.
Optical based drives, for example:- CD drive (ROM and RW) DVD drive (RPM and RW)
Flash or solid state chip based drives, for example:- USB drive. SD cards.
Outlines because you would not want the audience to know what exactly is on the slides, but just an outline to give them an idea of what is to come.
