I read the solution to this question, which seems to be the same target. My solution is (I'm sure) not the quickest, but some test case is apparently taking longer than 4s. I did a test case with 10,000 random integers and it completed in just over .33 seconds, so I'm trying to find what would result in this code hanging.
Here are the requirements:
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.
For a = [2, 1, 3, 5, 3, 2], the output should be firstDuplicate(a) = 3.
There are 2 duplicates: numbers 2 and 3. The second occurrence of 3 has a smaller index than the second occurrence of 2 does, so the answer is 3.
For a = [2, 4, 3, 5, 1], the output should be firstDuplicate(a) = -1.
def firstDuplicate(a): currentdupe = -1 dist = len(a) if len(a) == 0: return -1 if len(a) == 1: return -1 if len(a) == 2 and a == a: return a else: #new list b = list(a) b.sort() #check each double in the sorted range for x in (range(len(b)-1)): if b[x] == b[x+1]: #if distance is less than last found, use this one if a[a.index(b[x])+1:].index(b[x]) < dist: dist = a[a.index(b[x])+1:].index(b[x]) currentdupe = b[x] return currentdupe
Even in a worst case of 9,998 values of '1' and 2 values of '2', it still ran in just over .5s. When it runs a test case I get "Program exceeded the 4s time limit." How can this be?