Suppose we are performing a binary search on a sorted array called numbers initialized as follows: // index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 int[] numbers = {-5, -1, 0, 3, 9, 14, 19, 24, 33, 41, 56, 62, 70, 88, 99}; int index = binarySearch(numbers, 18); Write the indexes of the elements that would be examined by the binary search (the mid values in our algorithm's code).

The indexes of the elements that would be examined by the binary search are

7119

numbers[7] = 39

numbers[11] = 57

numbers[9] = 42

The values that would be returned from the search are    

395742

Explanation:

The complexity of searching a value in an array using binary search is O (log n). It follows divide and conquer principle. First we have to sort the elements in the array. Here in our case the array is already sorted.

   target = (search for the value) =42

numbers[] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14    

min max

here assign min= 0 (minimum index)

max= 14 (maximum index)    

Instead of searching for the target in a sequential manner we are searching by examining the middle term in the array by using the following formula. middle = ( min + max )/2

step 1) middle = (0 + 14)/2 = 7 numbers[middle]=numbers[7] = 39

compare target value with numbers[middle]

i.e target = 42 > 39 , the target value is greater than the numbers[middle]. so we have to move to upper part of the array.

Then min= middle+1 = 7+1 = 8

max= (unchanged) 14

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14    

min max

step 2) middle = (8+ 14)/2 = 11 numbers[middle]=numbers[11] = 57

compare target value with numbers[middle]

i.e target =  42 < 57 ,the target value is lesser than the numbers[middle] .

Then min= (unchanged) 8

max= middle -1 =11-1 =10

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14    

min max

step 3) middle = (8+10)/2 = 9   numbers[middle]=numbers[9] = 42

i.e target =  42 = 42

Here stop the process. In this way we found our target using binary search.

this would definitely be a.

explanation: