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;
}
Answer:
while (quantity >= 50)
Explanation:
Required
Stop when quantity is less than 50
To do this, we make use of a while statement and the syntax is:
while (condition){ }
If the loop should stop when quantity is less than 50; then it means the loop would continue when quantity is greater or equal to 50
So, we have:
while (quantity >= 50)
All server side errors start with 5XX
