Answer:
The main benefit of the ordered list is that you can apply Binary Search( O( n log n) ) to search the elements. Instead of an unordered list, you need to go through the entire list to do the search( O(n) ).
The main cost of the ordered list is that every time you insert into a sorted list, you need to do comparisons to find where to place the element( O( n log n) ). But, every time you insert into an unsorted, you don't need to find where to place the element in the list ( O(1) ). Another cost for an ordered list is where you need to delete an element, you have an extra cost rearranging the list to maintain the order.
Answer:
email is the easier way of communicating and fast
Answer:
Option D
Explanation:
option d component level design model is the correct answer
// Variable to keep track of array size
int length = 0;
// Array itself
int array[] = {};
// while loop will take input in the array until a negative number is entered
while(input>=0){
stdin = new Scanner(System.in);
array[length] = stdin;
length +=1;
}
// int variable to terminate while loop
int i =0;
// keep track of index of output array
int y =0;
while(arr[i] != length){
// making output array
int output[]={}; bool flag;
// put the element in out put array considering if it the desired one
output[y] = array [i];
// Now check if it was the desired?
for(int z=1;z<=length;z++){
if(array[i+z]!=output[y]){
flag = true;
}
else
{
output[y]=0;
y+=1;
}
if(array[i+z+1]=output[y] && flag == true){
output[y] = array[i];
y+=1;
}
}
//output the array
for(int o=0;o<y;o++)
System.out.println(output[o];
