LikelyWord | Topcoder Archive

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

```
{"time","zoology"}
```
```
1
```
```
Returns: 0
```
There are many more 1 letter words before "time" than either between "time" and "zoology" or after "zoology".
```
{"az","ma","xz"}
```
```
1
```
```
Returns: 1
```
12 words ("b", "c", ..., "m") would have index 1, while 11 ("n", ... , "x") would have index 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".
```
{"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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