Here is an array with exactly 15 elements:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Suppose that we are doing a binary search for an element. Indicate any elements that will be found by examining two or fewer numbers from the array.
What I've got: as we are doing binary search, so the number found by only one comparison will be 7th element = 7. For two comparison, this leads to second division of array. That is, number found can be either 3 or 11.
Am I right or not?
You are almost right, the first number is not seven but eight.
The others 2 will then be 4 and 12.
The correct answer would be 4, 8, 12
`I found the answer to be 8 that is the 7th element, the other elements found were 3.5th and 10.5th element of the sorted array. So, the next two numbers delved are 4 and 11.
explanation on how i got the answers.
given array is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
head=1
tail=15
middle= 0+14/2=7th element **0 is the index value of 1 and 14 is of 15**
middle value turns to be 8 as it is the 7th element.
solving value for first half
head=1
tail=8
middle= 0+7/2=3.5 or 3rd element **0 is the index value of 1 and 7 is of 8**
middle value now turns to be 4 as it is the 3rd element.
solving value for second half
head=8
tail=15
middle= 7+14/2=10.5 or 10th element **7 is the index value of 8 and 14 is
of 15**
middle value now turns to be 11 as it is the 10th element of the array`
Blockquote
4 8 12.arr[7] = 8, which may be leading to your confusion. You said the 7th element, you might mean the element at7.