Commit 7a30307

@@ -2,8 +2,8 @@
 Note: Write a solution with O(n) time complexity and O(1) additional space complexity, 
 since this is what you would be asked to do during a real interview.
 
-Given an array a that contains only numbers in the range from 1 to a.length, 
-find the first duplicate number for which the second occurrence has the minimal index. 
+Given an array a that contains only numbers in the range from 1 to a.length,
+find the first duplicate number for which the second occurrence has the minimal index.
 In other words, if there are more than 1 duplicated numbers, 
 return the number for which the second occurrence has a smaller index than the second occurrence of the other number does. 
 If there are no such elements, return -1.
@@ -36,19 +36,24 @@ DOC
 
 describe "first_duplicate" do
   def first_duplicate(items)
-    head, tail = 0, items.size - 1
+    head, tail = 0, 1
     min = -1
+    n = items.size
 
-    until head == (items.size - 1) || head == min
+    until head == (n - 1) || head == min
+      puts [items[head], items[tail]].inspect
       if items[head] == items[tail]
         min = min < 0 ?  tail + 1 : [min, tail + 1].min
+        puts ['match', items[head], items[tail], min].inspect
         head += 1
+        tail = head + 1
         break if tail == head
       else
-        tail -= 1
-        if tail == head
-          tail = items.size - 1
+        tail += 1
+        break if min >= 0 && tail > min
+        if tail == n
           head += 1
+          tail = head + 1
         end
       end
     end