Answer:
My goal is to have good grades because I failing now and I want to try and try to get my grade up if I do I will be successful and when I grow up I will have a good job and a better life for me and my family.If I success my life will be so different.
Answer:
To support software products, developers update software code to fix bugs, improve functionality, and even add new features. These changes are collected and released in updates, patches, and new product versions that can be installed by users
Answer:
DNS(Domain Name Services) is the correct answer.
Explanation:
DNS is the type of directory system which is distributed that finds the solution of a hostname which is readable by the human being. It is also critical in centralizing control of the confirmation and policies of the security in the environment of the domain with Active Directory. That's why the following answer is correct.
Answer: D) Service Level Agreement.
Explanation: A service Level Agreement, also known by the acronym SLA, is a written agreement between a service provider and its customer in order to set the agreed level for the quality of such service. The SLA is a tool that helps both parties reach a consensus in terms of the level of service quality, in aspects such as response time, time availability, available documentation, personnel assigned to the service, among others.
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;
}