Answer:
false.
Explanation:
it was introduced in 1994
I think .ppt
If I’m wrong I’m sorry
Ik lil bit but this is my max for now
Answer:
A private cloud is a set of applications and resources that are available only to individuals within a specific organization. These resources may be owned by the organization itself or by a third party, and may be hosted inside or outside
Explanation:
.
Answer:
Follows are the solution to this question:
Explanation:
These functions to interpret but instead, print their for store assuming that have such a healthy connection, which data structure dictionary promoting each searching operation, add, remove, minimum, maximum, successor, and O(log n) time was its pre successor.
- Use its dictionary only for following abstract procedures, minimal, succeeding, in O(n login) time, search, add. It uses a dictionary with only the corresponding conceptual functions, minimize, add, in O(n log n) time, remove, search.
- Use the dictionary only for corresponding abstract procedures to sort, add, and sort in O(n log n) time. and in-order traversal. It is the most and minimum shop value, that want to be checked and updated from each deletion.
- When a minimum is removed, call the holder, change the minimum, and remove the single object. Once inserted, evaluate the brand new item to min/max, then update min/max if it replaces either one.
Answer:
You can simplify the problem down by recognizing that you just need to keep track of the integers you've seen in array that your given. You also need to account for edge cases for when the array is empty or the value you get would be greater than your max allowed value. Finally, you need to ensure O(n) complexity, you can't keep looping for every value you come across. This is where the boolean array comes in handy. See below -
public static int solution(int[] A)
{
int min = 1;
int max = 100000;
boolean[] vals = new boolean[max+1];
if(A.length == 0)
return min;
//mark the vals array with the integers we have seen in the A[]
for(int i = 0; i < A.length; i++)
{
if(A[i] < max + 1)
vals[A[i]] = true;
}
//start at our min val and loop until we come across a value we have not seen in A[]
for (int i = 1; i < max; i++)
{
if(vals[i] && min == i)
min++;
else if(!vals[i])
break;
}
if(min > max)
return max;
return min;
}
