Fix word split scoring: prefer longer words over short ones

"Comeon" was split as "Com eon" instead of "Come on" because both
are 2-word splits. Now uses sum-of-squared-lengths as tiebreaker:
"come"(16) + "on"(4) = 20 > "com"(9) + "eon"(9) = 18.

Co-Authored-By: Claude Opus 4.6 (1M context) <noreply@anthropic.com>

klausur-service/backend/cv_review.py

@@ -735,36 +735,39 @@ def _try_split_merged_word(token: str) -> Optional[str]:
     lower = token.lower()
     n = len(lower)
 
-    # dp[i] = shortest list of word lengths that covers lower[:i], or None
+    # dp[i] = (word_lengths_list, score) for best split of lower[:i], or None
+    # Score: (-word_count, sum_of_squared_lengths) — fewer words first,
+    # then prefer longer words (e.g. "come on" over "com eon")
     dp: list = [None] * (n + 1)
-    dp[0] = []
+    dp[0] = ([], 0)
 
     for i in range(1, n + 1):
-        # Try all possible last-word lengths (2..min(i, 20))
-        # Allow single-char words only for 'a' and 'I'
-        min_len = 1
         for j in range(max(0, i - 20), i):
             if dp[j] is None:
                 continue
-            word_len = i - j
             candidate = lower[j:i]
+            word_len = i - j
             if word_len == 1 and candidate not in ('a', 'i'):
                 continue
-            if word_len < 2 and candidate not in ('a', 'i'):
-                continue
             if _spell_dict_knows(candidate):
-                new_split = dp[j] + [word_len]
-                # Prefer fewer words (shorter split)
-                if dp[i] is None or len(new_split) < len(dp[i]):
-                    dp[i] = new_split
+                prev_words, prev_sq = dp[j]
+                new_words = prev_words + [word_len]
+                new_sq = prev_sq + word_len * word_len
+                new_key = (-len(new_words), new_sq)
+                if dp[i] is None:
+                    dp[i] = (new_words, new_sq)
+                else:
+                    old_key = (-len(dp[i][0]), dp[i][1])
+                    if new_key > old_key:
+                        dp[i] = (new_words, new_sq)
 
-    if dp[n] is None or len(dp[n]) < 2:
+    if dp[n] is None or len(dp[n][0]) < 2:
         return None
 
     # Reconstruct with original casing
     result = []
     pos = 0
-    for wlen in dp[n]:
+    for wlen in dp[n][0]:
         result.append(token[pos:pos + wlen])
         pos += wlen