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;
}
It means get in the kitchen woman that’s what cookery means
Answer:
Ada Lovelace was the first computer programmer
Answer:
A and C
Explanation:
Option A:
In IPv6 there is a rule to reduce an IPv6 address when there are two or more consecutive segments of zeros just one time. This rule says that you can change the consecutive zeros for “::”
Here is an example
How to reduce the following IPv6 address?
ff02:0000:0000:0000:0000:0000:0000:d500
Ans: ff02::d500
Example 2:
2001:ed02:0000:0000:cf14:0000:0000:de95
Incorrect Answer -> 2001:ed02::cf14::de95
Since the rule says that you can apply “::” just one time, you need to do it for a per of zero segments, so the correct answer is:
Correct Answer -> 2001:ed02::cf14:0:0:de95
Or
2001:ed02:0:0:cf14::de95
Option C:
Since in IPv6 there are 
available addresses which means 340.282.366.920.938.463.463.374.607.431.768.211.456 (too many addresses), there is no need of NAT solution, so each device can have its own IP address by the same interface to have access through the internet if needed. If not, you can block the access through internet by the firewall.
