JOIN

 Problem Statement

Problem Statement for LikelyWord

### Problem Statement

Given a dictionary of words, where will a newly coined word be most likely to fit?

We are given k, the length of the new word, and dictionary, a String[] of words in ascending alphabetical order. Suppose that the new word is equally likely to be any k-letter word that is not already in the dictionary. Return the most likely 0-based index for the new word. If there is more than one most likely index, return -1.

### Definition

 Class: LikelyWord Method: likely Parameters: String[], int Returns: int Method signature: int likely(String[] dictionary, int k) (be sure your method is public)

### Constraints

-dictionary will contain between 1 and 50 elements, inclusive.
-Each element of dictionary will contain between 1 and 50 characters, inclusive.
-Each character in each element of dictionary will be a lowercase letter ('a'-'z').
-The elements of dictionary will be distinct and in ascending alphabetical order.
-k will be between 1 and 10, inclusive.
-There will be at least one k-letter word that is not in dictionary.

### Examples

0)

 `{"time","zoology"}` `1`
`Returns: 0`
 There are many more 1 letter words before "time" than either between "time" and "zoology" or after "zoology".
1)

 `{"az","ma","xz"}` `1`
`Returns: 1`
 12 words ("b", "c", ..., "m") would have index 1, while 11 ("n", ... , "x") would have index 2.
2)

 `{"az","ma","xz"}` `2`
`Returns: 2`
 With the same dictionary but a longer new word, it becomes most likely that the new word will go between "ma" and "xz".
3)

 `{"a","m","y"}` `1`
`Returns: -1`
 There are 23 equally likely 1-letter words (since 3 are already in the dictionary). 0 would have index 0, 11 would have index 1, 11 would have index 2, and 1 would have index 3. So no index is most likely.

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This problem was used for:
Single Round Match 336 Round 1 - Division I, Level Two