39
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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, your program / function may result in undefined behaviour.

Example:

For a = [2, 3, 3, 1, 5, 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.

This is , so shortest answer in bytes wins.

BONUS: Can you solve it in O(n) time complexity and O(1) additional space complexity?

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Martin Ender Sep 3 '17 at 13:12

49 Answers 49

15
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Python 2, 34 bytes

O(n2) time, O(n) space

Saved 3 bytes thanks to @vaultah, and 3 more from @xnor!

lambda l:l[map(l.remove,set(l))<0]

Try it online!

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  • 1
    \$\begingroup\$ lambda l:map(l.remove,set(l))and l[0] is shorter. \$\endgroup\$ – vaultah Jul 30 '17 at 21:29
  • 1
    \$\begingroup\$ It looks like lambda l:l[map(l.remove,set(l))<0] works, even though the order of evaluation is weird. \$\endgroup\$ – xnor Jul 31 '17 at 0:08
  • \$\begingroup\$ This doesn't return -1 when no duplicates are found without the 'footer code', does that code not count towards the bytes? I'm new to code golf, sorry if it's a basic question! \$\endgroup\$ – Chris_Rands Jul 31 '17 at 12:38
  • \$\begingroup\$ @Chris_Rands Beneath the question musicman did ask if exception is okay instead of -1 and OP said its okay and musicman's answer throws exception. \$\endgroup\$ – LiefdeWen Jul 31 '17 at 13:55
  • \$\begingroup\$ That took me a while to figure out. Well played. Getting the 0th element of l using the conditional after modifying it is really clever. \$\endgroup\$ – Thoth19 Jul 31 '17 at 22:45
11
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JavaScript (ES6), 47 36 31 25 bytes

Saved 6 bytes thanks to ThePirateBay

Returns undefined if no solution exists.

Time complexity: O(n) :-)
Space complexity: O(n) :-(

a=>a.find(c=>!(a[-c]^=1))

How?

We keep track of already encountered values by saving them as new properties of the original array a by using negative numbers. This way, they can't possibly interfere with the original entries.

Demo

let f =

a=>a.find(c=>!(a[-c]^=1))

console.log(f([2, 3, 3, 1, 5, 2]))
console.log(f([2, 4, 3, 5, 1]))
console.log(f([1, 2, 3, 4, 1]))

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  • \$\begingroup\$ 25 bytes: a=>a.find(c=>!(a[-c]^=1)) \$\endgroup\$ – user72349 Jul 30 '17 at 22:11
  • \$\begingroup\$ @ThePirateBay Oh, of course. Thanks! \$\endgroup\$ – Arnauld Jul 30 '17 at 22:14
  • \$\begingroup\$ Just notice that Objects in JavaScript may not be implemented as hash table. Time complexity of accessing keys of some object may not be O(1). \$\endgroup\$ – tsh Aug 1 '17 at 9:47
6
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Mathematica, 24 bytes

#/.{h=___,a_,h,a_,h}:>a&

Mathematica's pattern matching capability is so cool!

Returns the original List for invalid input.

Explanation

#/.

In the input, replace...

{h=___,a_,h,a_,h}

A List with a duplicate element, with 0 or more elements before, between, and after the duplicates...

... :>a

With the duplicate element.

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6
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Jelly, 5 bytes

Ṛœ-QṪ

Try it online!

How it works

Ṛœ-QṪ  Main link. Argument: A (array)

Ṛ      Yield A, reversed.
   Q   Unique; yield A, deduplicated.
 œ-    Perform multiset subtraction.
       This removes the rightmost occurrence of each unique element from reversed
       A, which corresponds to the leftmost occurrence in A.
    Ṫ  Take; take the rightmost remaining element, i.e., the first duplicate of A.
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  • \$\begingroup\$ œ- removes the rightmost occurrences? TIL \$\endgroup\$ – Erik the Outgolfer Jul 31 '17 at 9:42
  • \$\begingroup\$ This doesn't seem to return -1 for no duplicates. Throwing an exception is okay as per OP but I'm not sure if 0 is even though it's not in the range. \$\endgroup\$ – Erik the Outgolfer Jul 31 '17 at 12:18
5
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Haskell, 35 bytes

f s(h:t)|h`elem`s=h|1<2=f(h:s)t
f[]

Try it online! Crashes if no duplicate is found.

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4
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Jelly, 6 bytes

xŒQ¬$Ḣ

Try it online!

Returns the first duplicate, or 0 if there is no duplicate.

Explanation

xŒQ¬$Ḣ  Input: array M
    $   Operate on M
 ŒQ       Distinct sieve - Returns a boolean mask where an index is truthy
          for the first occurrence of an element
   ¬      Logical NOT
x       Copy each value in M that many times
     Ḣ  Head
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  • \$\begingroup\$ It's golfier to use indexing like this: ŒQi0ị. \$\endgroup\$ – Erik the Outgolfer Jul 31 '17 at 12:23
  • \$\begingroup\$ @EriktheOutgolfer If there are no duplicates, i0 would return 0, where would index and return the last value of the input instead of 0. \$\endgroup\$ – miles Jul 31 '17 at 12:31
4
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Japt, 7 bytes

æ@bX ¦Y

Test it online!

Explanation

 æ@   bX ¦ Y
UæXY{UbX !=Y}  Ungolfed
               Implicit: U = input array
UæXY{       }  Return the first item X (at index Y) in U where
     UbX         the first index of X in U
         !=Y     is not equal to Y.
               In other words, find the first item which has already occured.
               Implicit: output result of last expression

Alternatively:

æ@¯Y øX

Test it online!

Explanation

 æ@   ¯ Y øX
UæXY{Us0Y øX}  Ungolfed
               Implicit: U = input array
UæXY{       }  Return the first item X (at index Y) in U where
     Us0Y        the first Y items of U (literally U.slice(0, Y))
          øX     contains X.
               In other words, find the first item which has already occured.
               Implicit: output result of last expression
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4
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Pyth, 5 bytes

h.-Q{

Test suite

Remove from Q the first appearance of every element in Q, then return the first element.

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  • \$\begingroup\$ @LuisMendo Ok thanks. Sorry for creating confusion, I should learn to read... \$\endgroup\$ – Mr. Xcoder Jul 31 '17 at 11:53
  • \$\begingroup\$ @Mr.Xcoder No, it's the OP's fault. That information should be in the challenge text, but just in a comment \$\endgroup\$ – Luis Mendo Jul 31 '17 at 12:40
4
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Dyalog APL, 27 24 20 19 13 12 11 bytes

⊢⊃⍨0⍳⍨⊢=⍴↑∪

Now modified to not depend on v16! Try it online!

How? (With input N)

  • ⊢⊃⍨... - N at this index:
    • ⍴↑∪ - N with duplicates removed, right-padded with 0 to fit N
    • ⊢= - Element-wise equality with N
    • 0⍳⍨ - Index of the first 0. `
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  • \$\begingroup\$ nevermind, I misread the question. not enough test cases though... \$\endgroup\$ – Uriel Jul 30 '17 at 21:44
  • \$\begingroup\$ Sorry for misleading you, I also misread the question. \$\endgroup\$ – miles Jul 30 '17 at 22:12
  • \$\begingroup\$ Looks like 36 bytes to me. \$\endgroup\$ – Adám Jul 31 '17 at 11:09
  • \$\begingroup\$ Oh god, iota underbar isn't in ⎕AV, is it? \$\endgroup\$ – Zacharý Jul 31 '17 at 11:22
  • \$\begingroup\$ @Zacharý Right, Classic translates it to ⎕U2378  when loading. Try it online! \$\endgroup\$ – Adám Aug 4 '17 at 4:44
3
+100
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Python 3, 94 92 bytes

O(n) time and O(1) extra memory.

def f(a):
 r=-1
 for i in range(len(a)):t=abs(a[i])-1;r=[r,i+1][a[t]<0>r];a[t]*=-1
 return r

Try it online!

Source of the algorithm.

Explanation

The basic idea of the algorithm is to run through each element from left to right, keep track of the numbers that have appeared, and returning the number upon reaching a number that has already appeared, and return -1 after traversing each element.

However, it uses a clever way to store the numbers that have appeared without using extra memory: to store them as the sign of the element indexed by the number. For example, I can represent the fact that 2 and 3 has already appeared by having a[2] and a[3] negative, if the array is 1-indexed.

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  • \$\begingroup\$ What would this do for i where a[i] > n? \$\endgroup\$ – Downgoat Jul 31 '17 at 6:02
  • \$\begingroup\$ @Downgoat read the question again. \$\endgroup\$ – Leaky Nun Jul 31 '17 at 6:04
  • \$\begingroup\$ The question says 1 to a.length but for a[i]= a.length wouldn't this go out of bounds? \$\endgroup\$ – Downgoat Jul 31 '17 at 6:05
  • \$\begingroup\$ @Downgoat t=abs(a[i])-1=a.length-1 \$\endgroup\$ – Leaky Nun Jul 31 '17 at 6:09
  • 3
    \$\begingroup\$ Note from feersum: "solution is cheating because it uses integers 1 bit larger than the input." \$\endgroup\$ – Leaky Nun Jul 31 '17 at 6:22
3
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Perl 6, 13 bytes

*.repeated[0]

Try it


Explanation

  • The * is in a Term position so the whole statement is a WhateverCode lambda.

  • The .repeated is a method that results in every value except for the first time each value was seen.

    say [2, 3, 3, 3, 1, 5, 2, 3].repeated.perl; # (3, 3, 2, 3).Seq
    #   (      3, 3,       2, 3).Seq
    
  • [0] just returns the first value in the Seq.
    If there is no value Nil is returned.
    (Nil is the base of the Failure types, and all types are their own undefined value, so Nil different than an undefined value in most other languages)


Note that since the implementation of .repeated generates a Seq that means it doesn't start doing any work until you ask for a value, and it only does enough work to generate what you ask for.
So it would be easy to argue this has at worst O(n) time complexity, and at best O(2) time complexity if the second value is a repeat of the first.
Similar can probably be said of memory complexity.

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3
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APL (Dyalog), 20 bytes

⊃n/⍨(,≢∪)¨,\n←⎕,2⍴¯1

Try it online!

2⍴¯1 negative one reshaped into a length-two list

⎕, get input (mnemonic: console box) and prepend to that

n← store that in n

,\ prefixes of n (lit. cumulative concatenation)

( apply the following tacit function to each prefix

, [is] the ravel (just ensures that the prefix is a list)

 different from

 the unique elements[?] (i.e. is does the prefix have duplicates?)

n/⍨ use that to filter n (removes all elements until the first for which a duplicate was found)

 pick the first element from that

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  • \$\begingroup\$ Wow, you got beat three times. Still, +1. And can you add an explanation of how this works? \$\endgroup\$ – Zacharý Aug 3 '17 at 22:10
  • \$\begingroup\$ @Zacharý Apparently I just needed to get the ball rolling. Here you go. \$\endgroup\$ – Adám Aug 4 '17 at 5:13
  • \$\begingroup\$ @Zacharý Eventually, I managed to beat them all. \$\endgroup\$ – Adám Aug 4 '17 at 8:09
3
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APL (Dyalog), 11 bytes

As per the new rules, throws an error if no duplicates exist.

⊢⊃⍨⍬⍴⍳∘≢~⍳⍨

Try it online!

⍳⍨ the indices of the first occurrence of each element

~ removed from

⍳∘≢ of all the indices

⍬⍴ reshape that into a scalar (gives zero if no data is available)

⊃⍨ use that to pick from (gives error on zero)

 the argument

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  • \$\begingroup\$ Well, yeah, when the rules are changed, of course you can beat them all! \$\endgroup\$ – Zacharý Aug 4 '17 at 20:05
  • \$\begingroup\$ Well, I tied you. \$\endgroup\$ – Zacharý Aug 4 '17 at 20:56
3
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APL, 15

{⊃⍵[(⍳⍴⍵)~⍵⍳⍵]}

Seems like we can return 0 instead of -1 when there are no duplicates, (thanks Adám for the comment). So 3 bytes less.

A bit of description:

⍵⍳⍵         search the argument in itself: returns for  each element the index of it's first occurrence
(⍳⍴⍵)~⍵⍳⍵   create a list of all indexes, remove those found in ⍵⍳⍵; i.e. remove all first elements
⊃⍵[...]     of all remaining elements, take the first. If the array is empty, APL returns zero

For reference, old solution added -1 to the list at the end, so if the list ended up empty, it would contain -1 instead and the first element would be -1.

{⊃⍵[(⍳⍴⍵)~⍵⍳⍵],¯1}

Try it on tryapl.org

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3
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Retina, 26 24 bytes

1!`\b(\d+)\b(?<=\b\1 .*)

Try it online! Explanation: \b(\d+)\b matches each number in turn, and then the lookbehind looks to see whether the number is a duplicate; if it is the 1st match is ! output, rather than the count of matches. Unfortunately putting the lookbehind first doesn't seem to work, otherwise it would save several bytes. Edit: Added 7 bytes to comply with the -1 return value on no match. Saved 2 bytes thanks to @MartinEnder.

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  • 2
    \$\begingroup\$ For the record, the lookaround won't backtrack. This prevents this from working if you try to put it before. I've made this mistake many times, and Martin always corrects me. \$\endgroup\$ – FryAmTheEggman Jul 31 '17 at 0:37
  • \$\begingroup\$ I got 30 bytes by using a lookahead instead of a lookbehind. Also, the rules now say you don't need to return -1. \$\endgroup\$ – Value Ink Aug 4 '17 at 7:45
  • \$\begingroup\$ @ValueInk But the correct answer for that test case is 3... \$\endgroup\$ – Neil Aug 4 '17 at 8:00
  • \$\begingroup\$ OH. I misread the challenge, whoops \$\endgroup\$ – Value Ink Aug 4 '17 at 8:47
2
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MATL, 8 bytes

&=Rsqf1)

Gives an error (without output) if no duplicate exists.

Try at MATL Online!

Explanation

&=   % Implict input. Matrix of all pairwise equality comparisons
R    % Keep the upper triangular part (i.e. set lower part to false)
s    % Sum of each column
q    % Subtract 1
f    % Indices of nonzero values
1)   % Get first. Gives an error is there is none. Implictly display
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2
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R, 34 bytes

c((x=scan())[duplicated(x)],-1)[1]

Cut a few characters off the answer from @djhurio, don't have enough reputation to comment though.

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  • \$\begingroup\$ oh...I didn't see this answer; this is good for the prior spec when missing values required -1 but with the new spec, I managed to golf it down even more. This is still solid and it's a different approach from the way he did it, so I'll give you a +1! \$\endgroup\$ – Giuseppe Jul 31 '17 at 14:22
2
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J, 17 16 bytes

(*/{_1,~i.&0)@~:

How?

(*/{_1,~i.&0)@~:

             @~: returns the nub sieve which is a vector with 1 for the first occurrence of an element in the argument and 0 otherwise

        i.&0     returns the first index of duplication

    _1,~         appends _1 to the index

 */              returns 0 with duplicates (product across nub sieve)

     {           select _1 if no duplicates, otherwise return the index
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2
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R, 28 bytes

(x=scan())[duplicated(x)][1]

Try it online!

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  • \$\begingroup\$ I think you can now return NA for missing values since the spec has changed; so (x=scan())[duplicated(x)][1] is perfectly valid. \$\endgroup\$ – Giuseppe Jul 31 '17 at 12:53
2
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J, 12 bytes

,&_1{~~:i.0:

Try it online!

Explanation

,&_1{~~:i.0:  Input: array M
      ~:      Nub-sieve
          0:  The constant 0
        i.    Find the index of the first occurrence of 0 (the first duplicate)
,&_1          Append -1 to M
    {~        Select the value from the previous at the index of the first duplicate
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2
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Dyalog APL Classic, 18 chars

Only works in ⎕IO←0.

     w[⊃(⍳∘≢~⍳⍨)w←¯1,⎕]

Remove from the list of indices of the elements of the argument with a prepended "-1" the list indices of its nub and then pick the first of what's left. If after the removal there only remains an empty vector, its first element is by definition 0 which is used to index the extended argument producing the desired -1.

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2
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Ruby, 28 36 bytes

Misunderstood the challenge the first time. O(n) time, O(n) space.

->a{d={};a.find{|e|b=d[e];d[e]=1;b}}

Try it online!

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2
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Java (OpenJDK 8), 65 117 109 bytes

Previous 65 byte solution:

r->{for(int a,b=0,z,i=0;;b=a)if((a=b|1<<(z=r[i++]))==b)return z;}

New solution. 19 bytes are included for import java.math.*;

-8 bytes thanks to @Nevay

r->{int z,i=0;for(BigInteger c=BigInteger.ZERO;c.min(c=c.setBit(z=r[i++]))!=c;);return z;}

Try it online!

Edit

The algorithm in my original program was fine, but the static size of the datatype used meant that it broke fairly quickly once the size went above a certain threshold.

I have changed the datatype used in the calculation to increase the memory limit of the program to accommodate this (using BigInteger for arbitrary precision instead of int or long). However, this makes it debatable whether or not this counts as O(1) space complexity.

I will leave my explanation below intact, but I wish to add that I now believe it is impossible to achieve O(1) space complexity without making some assumptions.

Proof

Define N as an integer such that 2 <= N .

Let S be a list representing a series of random integers [x{1}, ..., x{N}], where x{i} has the constraint 1 <= x{i} <= N.

The time complexity (in Big-O notation) required to iterate through this list exactly once per element is O(n)

The challenge given is to find the first duplicated value in the list. More specifically, we are searching for the first value in S that is a duplicate of a previous item on the list.

Let p and q be the positions of two elements in the list such that p < q and x{p} == x{q}. Our challenge becomes finding the smallest q that satisfies those conditions.

The obvious approach to this problem is to iterate through S and check if our x{i} exists in another list T: If x{i} does not exist in T, we store it in T. If x{i} does exist in T, it is the first duplicate value and therefore the smallest q, and as such we return it. This space efficiency is O(n).

In order to achieve O(1) space complexity while maintaining O(n) time complexity, we have to store unique information about each object in the list in a finite amount of space. Because of this, the only way any algorithm could perform at O(1) space complexity is if: 1. N is given an upper bound corresponding to the memory required to store the maximum number of possible values for a particular finite datatype. 2. The re-assignment of a single immutable variable is not counted against the complexity, only the number of variables (a list being multiple variables). 3. (Based on other answers) The list is (or at least, the elements of the list are) mutable, and the datatype of the list is preset as a signed integer, allowing for changes to be made to elements further in the list without using additional memory.

1 and 3 both require assumptions and specifications about the datatype, while 2 requires that only the number of variables be considered for the calculation of space complexity, rather than the size of those variables. If none of these assumptions are accepted, it would be impossible to achieve both O(n) time complexity and O(1) space complexity.

Explanation

Whoo boy, this one took an embarrassingly long time to think up a bit of brain power.

So, going for the bonus is difficult. We need both to operate over the entire list exactly once and track which values we've already iterated over without additional space complexity.

Bit manipulation solves those problems. We initialize our O(1) 'storage', a pair of integers, then iterate through the list, OR-ing the ith bit in our first integer and storing that result to the second.

For instance, if we have 1101, and we perform an OR operation with 10, we get 1111. If we do another OR with 10, we still have 1101.

Ergo, once we perform the OR operation and end up with the same number, we've found our duplicate. No duplicates in the array causes the program to run over and throw an exception.

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  • \$\begingroup\$ Also, your second test includes the number 100, but thats impossible since the array itself is only 5 long \$\endgroup\$ – SchoolBoy Aug 10 '17 at 14:54
  • \$\begingroup\$ Also, this fails since an int doesn't have enough storage. \$\endgroup\$ – SchoolBoy Aug 10 '17 at 14:59
  • \$\begingroup\$ @SchoolBoy Good catch. My only problem is that there doesn't seem to be any upper limit on the size of the array, so I can't realistically change my code to solve memory issues. \$\endgroup\$ – Xanderhall Aug 10 '17 at 16:26
  • \$\begingroup\$ @Xanderhall True, but i feel like 32 (or if you use a long, 64) numbers is too little :p. Either way, imposing a limit on the input, and then allocating the maximum memory needed and calling it O(1) memory is just a cheat. It is still O(n) since if the size of the input increased, so would this upper bound to the memory. Which is also why I think it is impossible to create an O(n) O(1) algorithm \$\endgroup\$ – SchoolBoy Aug 10 '17 at 16:29
  • \$\begingroup\$ @Xanderhall P.S. I'm getting close to your 65, I'm at 67 bytes :p \$\endgroup\$ – SchoolBoy Aug 10 '17 at 16:29
2
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PHP, 56 44 38 32 bytes

for(;!${$argv[++$x]}++;);echo$x;

Run like this:

php -nr 'for(;!${$argv[++$x]}++;);echo$x;' -- 2 3 3 1 5 2;echo
> 3

Explanation

for(
  ;
  !${                 // Loop until current value as a variable is truthy
    $argv[++$x]       // The item to check for is the next item from input
  }++;                // Post increment, the var is now truthy
);
echo $x;              // Echo the index of the duplicate.

Tweaks

  • Saved 12 bytes by using variables instead of an array
  • Saved 6 bytes by making use of the "undefined behavior" rule for when there is no match.
  • Saved 6 bytes by using post-increment instead of setting to 1 after each loop

Complexity

As can be seen from the commented version of the code, the time complexity is linear O(n). In terms of memory, a maximum of n+1 variables will be assigned. So that's O(n).

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  • \$\begingroup\$ Thanks for not using a weird encoding. But you should add the error_reporting option to the byte count (or use -n, which is free). \$\endgroup\$ – Titus Aug 11 '17 at 11:33
  • \$\begingroup\$ We've been here before. PHP notices and warnings are ignorable. I might as well pipe them to /dev/null, which is the same. \$\endgroup\$ – aross Aug 11 '17 at 13:02
  • \$\begingroup\$ I tend to remember the wrong comments. :) Isn´t this O(n)? \$\endgroup\$ – Titus Aug 11 '17 at 13:04
  • \$\begingroup\$ Yes it's linear \$\endgroup\$ – aross Aug 11 '17 at 13:05
  • \$\begingroup\$ How is that O(1) for additional space? You're literally assigning a new variable per n, which is O(n) \$\endgroup\$ – Xanderhall Aug 11 '17 at 14:46
2
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Java 8, 82 78 76 bytes No longer viable, 75 67 64 bytes below in edit

As a lambda function:

a->{Set<Long>s=new HashSet<>();for(long i:a)if(!s.add(i))return i;return-1;}

Probably can be made much smaller, this was very quick.

Explanation:

a->{                                //New lambda function with 'a' as input
    Set<Long>s=new HashSet<>();     //New set
    for(long i:a)                   //Iterate over a
        if(!s.add(i))               //If can't add to s, already exists
            return i;               //Return current value
        return-1;                   //No dupes, return -1
}

*Edit*

75 67 64 bytes using the negation strategy:

a->{int i=0,j;while((a[j=Math.abs(a[i++])-1]*=-1)<0);return++j;}

Try it online!

(-3 bytes thanks to @Nevay)

Explanation:

a->{                                         //New lambda expression with 'a' as input
    int i=0,j;                               //Initialise i and declare j
    while((a[j=Math.abs(a[i++])-1]*=-1)<0);  //Negate to keep track of current val until a negative is found
    return++j;                               //Return value
}

Loops over the array, negating to keep track. If no dupes, just runs over and throws an error.

Both of these work on O(n) time and O(n) space complexity.

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  • \$\begingroup\$ It's worth noting that this will need to be assigned to a lambda returning Number, since i is a long and -1 is an int. \$\endgroup\$ – Jakob Aug 10 '17 at 14:36
  • \$\begingroup\$ @Jakob Not necessary, -1 being an int will automatically be cast to a long without explicitly specifying the cast \$\endgroup\$ – SchoolBoy Aug 10 '17 at 14:47
  • \$\begingroup\$ It will cast implicitly to long, but not to Long as required for the lambda to be assigned to a Function. Did you test it? Regardless, that solution can be replaced with your new one. \$\endgroup\$ – Jakob Aug 10 '17 at 16:13
  • \$\begingroup\$ You can use raw types Set s=new HashSet(); to save 7 bytes. (Besides: afaik the import of java.util.*; has to be included into the byte count -> +19 bytes.) The return statement can be return++j, the if-statement can be removed a->{int i=0,j;for(;(a[j=Math.abs(a[i++])-1]*=-1)<0;);return++j;} (-3 bytes). \$\endgroup\$ – Nevay Aug 11 '17 at 13:34
2
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Brachylog, 5 bytes

a⊇=bh

Try it online!

Explanation

a⊇=bh  Input is a list.
a      There is an adfix (prefix or suffix) of the input
 ⊇     and a subsequence of that adfix
  =    whose elements are all equal.
   b   Drop its first element
    h  and output the first element of the rest.

The adfix built-in a lists first all prefixes in increasing order of length, then suffixes in decreasing order of length. Thus the output is produced by the shortest prefix that allows it, if any. If a prefix has no duplicates, the rest of the program fails for it, since every subsequence of equal elements has length 1, and the first element of its tail doesn't exist. If a prefix has a repeated element, we can choose the length-2 subsequence containing both, and the program returns the latter.

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  • \$\begingroup\$ Another 5 bytes solution: a⊇Ċ=h, which only looks at length-2 subsets. \$\endgroup\$ – Fatalize Dec 1 '17 at 7:20
1
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C#, 145 bytes

using System.Linq;a=>{var d=a.Where(n=>a.Count(t=>t==n)>1);return d.Select((n,i)=>new{n,i}).FirstOrDefault(o=>d.Take(o.i).Contains(o.n))?.n??-1;}

Probably a lot shorter way to do this in C# with a simple loop but I wanted to try it with Linq.

Try it online!

Full/Formatted version:

namespace System.Linq
{
    class P
    {
        static void Main()
        {
            Func<int[], int> f = a =>
            {
                var d = a.Where(n => a.Count(t => t == n) > 1);
                return d.Select((n, i) => new { n, i }).FirstOrDefault(o => d.Take(o.i).Contains(o.n))?.n ?? -1;
            };

            Console.WriteLine(f(new[] { 2, 3, 3, 1, 5, 2 }));
            Console.WriteLine(f(new[] { 2, 4, 3, 5, 1 }));

            Console.ReadLine();
        }
    }
}
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  • \$\begingroup\$ Here is the simple loop version. But I like the Linq version much more. \$\endgroup\$ – LiefdeWen Jul 31 '17 at 13:13
  • \$\begingroup\$ @LiefdeWen Post it as an answer :) Though I do usually like Linq better too :) Might be able to get it shorter with Linq too but I'm now sure. \$\endgroup\$ – TheLethalCoder Jul 31 '17 at 13:16
  • \$\begingroup\$ Nah, this question is overpopulated and I would rather you get the up-votes for this question. \$\endgroup\$ – LiefdeWen Jul 31 '17 at 13:17
1
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Haskell, 78 69 bytes

 fst.foldl(\(i,a)(j,x)->(last$i:[j|i<0,elem x a],x:a))(-1,[]).zip[1..]

Try it online!

Saved 9 bytes thanks to @nimi

A basic path through the list. If the current element has not yet been seen (i<0) and is in the accumulator list (elem x a) then store the current index. Else, keep the index -1. In any case, add the current element to the accumulator list.

EDIT: I did not read the question carefully enough: this code outputs the index of the second element of a duplicate element.

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  • \$\begingroup\$ You can use the "Shorter Conditional" from our "Tips for golfing in Haskell": \ ... ->(last$i:[j|i<0,elem x a],x:a). Also: no need for the f=, because unnamed functions are allowed. \$\endgroup\$ – nimi Jul 31 '17 at 14:44
  • \$\begingroup\$ @nimi thanks for the tip! \$\endgroup\$ – jferard Jul 31 '17 at 20:16
1
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Python 2, 71 65 bytes

Returns None if there is no duplicate element

Edit: -6 bytes thanks to @musicman523

def f(n):
 for a in n:
	u=-abs(a)
	if n[u]<0:return-u
	n[u]=-n[u]

Try it online!

O(n) time complexity, O(n) space complexity, O(1) auxiliary space.

As the input list uses O(n) space, the space complexity is bound by this. Meaning we cannot have a lower space complexity than O(n)

Does modify the original list, if this is not allowed we could do it in the same complexity with 129 bytes

Explanation

Since every element is greater than 0 and less than or equal to the size of the list, the list has for each element a, an element on index a - 1 (0 indexed). We exploit this by saying that if the element at index i is negative, we have seen it before.

For each element a in the list n, we let u be negative the absolute value of a. (We let it be negative since python can index lists with negative indices, and we would otherwise need to do u=abs(a)-1) If the element at index u in the list is negative, we have seen it before and can therefore return -u (to get the absolute value of a, as all elements are positive). Else we set the element at index u to be negative, to remember that we have seen an element of value a before.

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  • \$\begingroup\$ Nice job! 65 bytes \$\endgroup\$ – musicman523 Aug 10 '17 at 5:50
  • \$\begingroup\$ Are you sure this is O(1) in memory? You are still using n bits of memory to store what numbers have already been visited, even though the bits are in the sign. It seems to me that this is O(n) in disguise \$\endgroup\$ – Sriotchilism O'Zaic Aug 10 '17 at 7:06
  • \$\begingroup\$ Technically this uses O(n) space - the n sign bits. If the array can only hold values between 1 and n, like how it was given, then it obviously doesn't work. \$\endgroup\$ – Oliver Ni Aug 10 '17 at 7:36
  • \$\begingroup\$ This really just comes down to the representation you choose for the numbers. If unsigned numbers are used, then this is O(n) auxiliary space. If signed numbers are used, then the sign bit is already there, meaning O(1) auxiliary space. \$\endgroup\$ – Halvard Hummel Aug 10 '17 at 7:52
  • \$\begingroup\$ I agree with you there. I personally would let you slide using signed integers as long as you didn't use the sign bit, it should be about the algorithm not the technicalities of the system. That being said I do think if you are going to use the sign bits you have to count them. I think this answer is pretty clever. If I any votes left today I would upvote it to counteract the downvote. \$\endgroup\$ – Sriotchilism O'Zaic Aug 10 '17 at 14:32
1
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Jelly, 4 bytes

ŒQi0

Try it online!

In case that all elements are unique, this returns 0 (undefined behavior).

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