# My solution using a Bloom filter - O(n) time, O(1) memory

• Ok this may not be the most efficient in the case of an array of int (the real gain would be for a stream of int or strings), but I thought it could be a good way to try this concept.

O(1) memory is actually pretty high here.

public class Solution {
int hashes=5;
int bits=5000000;
int[] primes = {31, 37,  41,  43, 47, 53, 59, 61, 67, 71};

public int findDuplicate(int[] nums) {

for(int i=0; i<nums.length; i++){
int count=0;
for(int h=0; h<hashes; h++){
String s =Integer.toString(nums[i]);
int hash = 0;
for(int c=0; c<s.length();c++)
hash = primes[h]*hash + s.charAt(c);
hash =  hash % bits;
if(hash<0)
hash = ( bits + hash) %bits;
count++;
}
if (count==0)
return nums[i];
}
return -1;
}
}

• Definitely the best solution for industrial application.

• This post is deleted!

• Wrong solution. First of all, bloom filter is typically used to check non-existence of element; second, even if say we can tolerate some small false positive rate, that would require O(N) space.

Your solution may fool leetcode OJ, but not a qualified interviewer.

• the best solution for an industrial application is to simply have a hashtable, and insert each element one at a time into the table. If the element is already there, return that element. However, that's O(n) space.

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