Answer:
hakkuna matata is the radious of a circle
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;
}
The answer is true because I don't know
Answer:
The answer is "Option b".
Explanation:
In the given code, a static method "inCommon" is declared, that accepts two array lists in its parameter, and inside the method two for loop is used, in which a conditional statement used, that checks element of array list a is equal to the element of array list b. If the condition is true it will return the value true, and if the condition is not true, it will return a false value. In this, the second loop is not used because j>0 so will never check the element of the first element.
