Byte Pair Encoding (BPE) Merges

GPT-style tokenizers use Byte Pair Encoding: start from single characters and repeatedly merge the most common pair of neighbors into a new symbol.

Split every word in words into characters. Repeated words count multiple times.
Count every pair of adjacent symbols across all words.
Pick the most frequent pair. If pairs tie, pick the one that is smallest when you compare (first symbol, second symbol) as strings.
Replace every occurrence of that pair, left to right, with the merged symbol, and record the merged symbol.
Repeat num_merges times, or stop early when no word has two symbols left.

Implement bpe_merges(words, num_merges).

Input

words: The training words (lowercase letters).

num_merges: How many merges to learn.

Output

Return the merged symbols in the order they were learned.

Constraints

1 ≤ number of words ≤ 1,000, 1 ≤ length of each word ≤ 30

0 ≤ num_merges ≤ 200

Examples

Example 1

Input:
words = ["low", "lower", "lowest"]
num_merges = 2

Expected output:
["lo", "low"]

Explanation

Pairs: "l o" and "o w" appear 3 times each; the tie goes to "l o" ("l" < "o"), giving "lo". Now "lo w" appears 3 times, giving "low". Result: ["lo", "low"].

Example 2

Input:
words = ["aaa"]
num_merges = 3

Expected output:
["aa", "aaa"]

Explanation

"aaa" has the pair "a a" twice. Merging left to right gives "aa" + "a". Then "aa a" merges into "aaa", and the word is a single symbol, so the third merge has nothing to do. Result: ["aa", "aaa"].

References

Run the sample tests to check your code without submitting.

Example 1 Not run

Input:

words = ["low", "lower", "lowest"]
num_merges = 2

Expected output:

["lo", "low"]

Your output

Example 2 Not run
Input:
```
words = ["aaa"]
num_merges = 3
```
Expected output:
```
["aa", "aaa"]
```
Your output

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1. Declaring Variables and Constants

Java requires you to declare the type of each variable. Use final for constants.

int age = 30;
double pi = 3.14159;
char grade = 'A';
String name = "Alice";
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