CODEWITHSUNDEEP
javaarraysalgorithmrepeat
AMH9
AMH9

Reputation: 179

repeated element in Array

I have an array of N elements and contain 1 to (N-1) integers-a sequence of integers starting from 1 to the max number N-1-, meaning that there is only one number is repeated, and I want to write an algorithm that return this repeated element, I have found a solution but it only could work if the array is sorted, which is may not be the case. ? 

int i=0;
while(i<A[i])
{
i++
}
int rep = A[i];

Upvotes: 0

Views: 177

Answers (3)

vlad_tepesch
vlad_tepesch

Reputation: 6891

I do not know why RC removed his comment but his idea was good.

With the knowledge of N you easy can calculate that the sum of [1:N-1]. then sum up all elementes in your array and subtract the above sum and you have your number.

This comes at the cost of O(n) and is not beatable.

However this only works with the preconditions you mentioned.

A more generic approach would be to sort the array and then simply walk through it. This would be O(n log(n)) and still better than your O(n²).

I you know the maximum number you may create a lookup table and init it with all zeros, walk through the array and check for one and mark the entries with one. The complexity is also just O(n) but at the expense of memory.

if the value range is unknown a simiar approach can be used but instead of using a lookup table a hashset canbe used.

Upvotes: 1

Ivaylo Strandjev
Ivaylo Strandjev

Reputation: 70929

A possible solution is to sum all elements in the array and then to compute the sym of the integers up to N-1. After that subtract the two values and voila - you found your number. This is the solution proposed by vlad_tepesch and it is good, but has a drawback - you may overflow the integer type. To avoid this you can use 64 bit integer.

However I want to propose a slight modification - compute the xor sum of the integers up to N-1(that is compute 1^2^3^...(N-1)) and compute the xor sum of your array(i.e. a0^a1^...aN-1). After that xor the two values and the result will be the repeated element.

Upvotes: 0

ursa
ursa

Reputation: 4591

Linear search will help you with complexity O(n):

final int n = ...;
final int a[] = createInput(n); // Expect each a[i] < n && a[i] >= 0
final int b[] = new int[n];

for (int i = 0; i < n; i++)
    b[i]++;

for (int i = 0; i < n; i++)
   if (b[i] >= 2)
      return a[i];

throw new IllegalArgumentException("No duplicates found");

Upvotes: 0

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