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Given a properly prepared String array containing five elements, which range of results could a proper invocation of Arrays.binarySearch() produce?

A. 0 through 4 B. 0 through 5 C. -1 through 4 D. -1 through 5 E. -5 through 4 F. -5 through 5 G. -6 through 4 H. -6 through 5

And the answer is given as G. As per my understanding the binarysearch index would return -1 if the match is not found since 0 is also considered as a successful match. So for 0 it is -1, 1 it is -2 and so on until 4 it is -5. So as per my understanding the answer would be E. Isn't it? Please clarify.

If the element used in binarySearch is greater than all elements in the array, it would have to be inserted at position 5. Therefore binarySearch returns -(5) - 1 = -6 in this case. [ July 24, 2008: Message edited by: Ralph Jaus ]

Unsuccessful searches return an int index that represents the insertion point. The insertion point is the place in the collection/array where the element would be inserted to keep the collection/array properly sorted. Because posi- We�ve talked a lot about sorting by natural order and using Comparators to sort. The last rule you�ll need to burn in is that, whenever you want to sort an array or a collection, the elements inside must all be mutually comparable. In other words, if you have an Object[] and you put Cat and Dog objects into it, you won�t be able to sort it. In general, objects of different types should be considered NOT mutually comparable, unless specifically stated otherwise. tive return values and 0 indicate successful searches, the binarySearch() method uses negative numbers to indicate insertion points. Since 0 is a valid result for a successful search, the first available insertion point is -1. Therefore, the actual insertion point is represented as (-(insertion point) -1). For instance, if the insertion point of a search is at element 2, the actual insertion point returned will be -3.

[ July 24, 2008: Message edited by: Raphael Rabadan ]

Thanks Ralph and Raphael for your reply... But this is where i am getting stuck.

"Since 0 is a valid result for a successful search, the first available insertion point is -1. Therefore,the actual insertion point is represented as (-(insertion point) -1). Forinstance, if the insertion point of a search is at element 2, the actual insertion point returned will be -3."

So the insertion point for the 5th element (4th in index) is going to be -5 right, how did we get to the -6, is that means say i have an array of 10 elements (0 to 9), the values would be from -11 through 9, is that the answer for the question then.....

Ralph Jaus
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Joined: Apr 27, 2008
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Given a = {0,1,2,3,4}. Because the array doesn't contain 5, Arrays.binarySearch(a,5) looks at what position the 5 is to be inserted to keep the array sorted. And that's the position with index 5. Therefore the result is -(5) -1 = -6. [ July 24, 2008: Message edited by: Ralph Jaus ]

Justin Smith
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Joined: Jul 24, 2008
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Thanks Ralph now i got the point So if the size of array is N the binarysearch will go through from

-(N+1) through (N-1)

Thanks again for the help

Raphael Rabadan
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Joined: Jul 05, 2008
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Try this code:

Ralph Jaus
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Joined: Apr 27, 2008
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Hi Justin, you're right.

By the way, the satements

No insert Point --> -1 At 0 --> -2 At 1 --> -3 At 2 --> -4 At 3 --> -5

are wrong. -1 is returned, if the the element used in binarySearch is less than the smallest element in the array (that is, it has to be inserted at position 0 to keep the array sorted). Let a = {0,2,4,6,8}. Than Arrays.binarySearch(a,1) would return -2, because 1 has to be inserted at index position 1 to keep the array sorted. And so on.

Oh, I see, Raphael has removed his wrong statements.

Raphael Rabadan
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Originally posted by Ralph Jaus: Oh, I see, Raphael has removed his wrong statements.

Yeah, to not confuse anyone :-)

Justin Smith
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Joined: Jul 24, 2008
Posts: 19

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Hi Ralph,

One more personal question, i saw the fantabulous score of yours (98%) and would like to know the secret behind the success. If you could share your preparation steps and notes that would be great.