Solving the “firstDuplicate” question in Python

I’m trying to solve the following challenge from

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.

The execution time limit is 4 seconds.

The guaranteed constraints were:

1 ≤ a.length ≤ 10^5, and

1 ≤ a[i] ≤ a.length

So my code was:

def firstDuplicate(a):
    b = a
    if len(list(set(a))) == len(a):
        return -1

    n = 0
    answer = -1
    starting_distance = float("inf")

    while n!=len(a):
        value = a[n]

        if a.count(value) > 1:

            place_of_first_number = a.index(value)

            a[place_of_first_number] = 'string'

            place_of_second_number = a.index(value)

            if place_of_second_number < starting_distance:

                starting_distance = place_of_second_number
                answer = value

        if n == len(a)-1:
            return answer 
    return answer

Out of the 22 tests the site had, I passed all of them up to #21, because the test list was large and the execution time exceeded 4 seconds. What are some tips for reducing the execution time, while keeping the the code more or less the same?


As @erip has pointed out in the comments, you can iterate through the list, add items to a set, and if the item is already in a set, it is a duplicate that has the lowest index, so you can simply return the item; or return -1 if you get to the end of the loop without finding a duplicate:

def firstDuplicate(a):
    seen = set()
    for i in a:
        if i in seen:
            return i
    return -1

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