beam: verify Floyd duplicate candidate occurs more than once

wizard-work/beam/find_duplicate.py

@@ -33,7 +33,12 @@ def find_duplicate(nums: List[int]) -> int:
     while slow2 != slow:
         slow2 = nums[slow2]
         slow = nums[slow]
-    return slow
+    candidate = slow
+    # Verify the candidate actually occurs more than once; a permutation
+    # (no duplicate) also has cycles, so Floyd alone is not sufficient.
+    if nums.count(candidate) < 2:
+        raise ValueError("no duplicate present in nums")
+    return candidate
 
 
 def _self_test() -> None: