Note

The first two approaches mentioned do not satisfy the constraints given in\nthe prompt, but they are solutions that you might be likely to come up with\nduring a technical interview. As an interviewer, I personally would not\nexpect someone to come up with the cycle detection solution unless they have\nheard it before.

Proof

Proving that at least one duplicate must exist in nums is simple\napplication of the\npigeonhole principle.\nHere, each number in nums is a "pigeon" and each distinct number that can\nappear in nums is a "pigeonhole". Because there are $$n+1$$ numbers are\n $$n$$ distinct possible numbers, the pigeonhole principle implies that at\nleast one of the numbers is duplicated.

Approach #1 Sorting [Accepted]

Intuition

If the numbers are sorted, then any duplicate numbers will be adjacent in the\nsorted array.

Algorithm

Given the intuition, the algorithm follows fairly simply. First, we sort the\narray, and then we compare each element to the previous element. Because\nthere is exactly one duplicated element in the array, we know that the array\nis of at least length 2, and we can return the duplicate element as soon as\nwe find it.

Complexity Analysis

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  Time complexity : $$O(nlgn)$$ \n

  \n

  The sort invocation costs $$O(nlgn)$$ time in Python and Java, so it\ndominates the subsequent linear scan.

  \n
\n
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  Space complexity : $$O(1)$$ (or $$O(n)$$ )

  \n

  Here, we sort nums in place, so the memory footprint is constant. If we\ncannot modify the input array, then we must allocate linear space for a\ncopy of nums and sort that instead.

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Approach #2 Set [Accepted]

Intuition

If we store each element as we iterate over the array, we can simply check\neach element as we iterate over the array.

Algorithm

In order to achieve linear time complexity, we need to be able to insert\nelements into a data structure (and look them up) in constant time. A Set\nsatisfies these constraints nicely, so we iterate over the array and insert\neach element into seen. Before inserting it, we check whether it is already\nthere. If it is, then we found our duplicate, so we return it.

Complexity Analysis

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  Time complexity : $$O(n)$$ \n

  \n

  Set in both Python and Java rely on underlying hash tables, so\ninsertion and lookup have amortized constant time complexities. The\nalgorithm is therefore linear, as it consists of a for loop that\nperforms constant work $$n$$ times.

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  Space complexity : $$O(n)$$ \n

  \n

  In the worst case, the duplicate element appears twice, with one of its\nappearances at array index $$n-1$$ . In this case, seen will contain\n $$n-1$$ distinct values, and will therefore occupy $$O(n)$$ space.

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Approach #3 Floyd\'s Tortoise and Hare (Cycle Detection) [Accepted]

Intuition

If we interpret nums such that for each pair of index $$i$$ and value\n $$v_i$$ , the "next" value $$v_j$$ is at index $$v_i$$ , we can reduce this\nproblem to cycle detection. See the solution to\nLinked List Cycle II\nfor more details.

Algorithm

First off, we can easily show that the constraints of the problem imply that\na cycle must exist. Because each number in nums is between $$1$$ and\n $$n$$ , it will necessarily point to an index that exists. Therefore, the list\ncan be traversed infinitely, which implies that there is a cycle.\nAdditionally, because $$0$$ cannot appear as a value in nums, nums[0]\ncannot be part of the cycle. Therefore, traversing the array in this manner\nfrom nums[0] is equivalent to traversing a cyclic linked list. Given this,\nthe problem can be solved just like\nLinked List Cycle II.

To see the algorithm in action, check out the animation below:

!?!../Documents/287_Find_the_Duplicate_Number.json:1280,720!?!

Complexity Analysis

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  Time complexity : $$O(n)$$ \n

  \n

  For detailed analysis, refer to \nLinked List Cycle II.

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  Space complexity : $$O(1)$$ \n

  \n

  For detailed analysis, refer to \nLinked List Cycle II.

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Analysis and solutions written by: @emptyset