Question

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).

Answer 1

Answer:

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

7 11 9

numbers[7] = 39

numbers[11] = 57

numbers[9] = 42

The values that would be returned from the search are    

39 57 42

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.