Question

Use numerals instead of words. If necessary, use / for the fraction bar.

Observe the given code.
private static int[] u={2,7,9,11,21,26,35,44};
public static int binaryS(int key)
{
int first=0;
int last= -1;
while(first<=last)
{
int middle=(first+last/2);
if (u[middle]==key)
return middle;
else if(u[middle]
first=middle+1;
else
last=middle-1;
}
return -1;
}
What will be stored in x after executing the following statement?
int x= binaryS(2);
The output will be :

Option 1: -1
Option 2: 0
Option 3: 1
Option 4: 2

Answer 1

The provided code has an error with the initialization of the 'last' variable. Assuming the code is corrected, executing the statement 'int x = binaryS(2);' would store 0 in x, as the number 2 is at the first position in the array 'u'.

Step-by-step explanation:

When looking at the code provided, we can see that it is an implementation of a binary search algorithm. However, the variable last is initially set to -1, which seems incorrect for a binary search, as it should generally be set to the length of the array minus one. This mistake will cause an infinite loop and thus the function will not work properly. Nonetheless, based on the intent of the code, when we call binaryS(2), we are attempting to find the position of the number 2 in the array u.

In a properly working binary search, the position of the number 2 would be at index 0 (since arrays are zero-indexed in Java). But since the last variable is set incorrectly, the while loop condition first <= last will never be true, and the function should directly return -1. However, if the last variable was set correctly (to the size of the array minus one), executing the function binaryS(2) would then return 0, because the number 2 is at the first position in the array.