# Find Duplicates of array using bit array

You have an array of N numbers, where N is at most 32,000. The array may have duplicates entries and you do not know what N is. With only 4 Kilobytes of memory available, how would print all duplicates elements in the array ?.

Examples:

```Input : arr[] = {1, 5, 1, 10, 12, 10}
Output : 1 10
1 and 10 appear more than once in given
array.

Input : arr[] = {50, 40, 50}
Output : 50
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We have 4 Kilobytes of memory which means we can address up to 8 * 4 * 210 bits. Note that 32 * 210 bits is greater than 32000. We can create a bit with 32000 bits, where each bit represents one integer.

Note: If you need to create a bit with more than 32000 bits then you can create easily more and more than 32000;

Using this bit vector, we can then iterate through the array, flagging each element v by setting bit v to 1. When we come across a duplicate element, we print it.

Below is Java implementation of the idea.

```// Java program to print all Duplicates in array
import java.util.*;
import java.lang.*;
import java.io.*;

// A class to represent array of bits using
// array of integers
class BitArray
{
int[] arr;

// Constructor
public BitArray(int n)
{
// Devide by 32. To store n bits, we need
// n/32 + 1 integers (Assuming int is stored
// using 32 bits)
arr = new int[(n>>5) + 1];
}

// Get value of a bit at given position
boolean get(int pos)
{
// Divide by 32 to find position of
// integer.
int index = (pos >> 5);

// Now find bit number in arr[index]
int bitNo  = (pos & 0x1F);

// Find value of given bit number in
// arr[index]
return (arr[index] & (1 << bitNo)) != 0;
}

// Sets a bit at given position
void set(int pos)
{
// Find index of bit position
int index = (pos >> 5);

// Set bit number in arr[index]
int bitNo = (pos & 0x1F);
arr[index] |= (1 << bitNo);
}

// Main function to print all Duplicates
static void checkDuplicates(int[] arr)
{
// create a bit with 32000 bits
BitArray ba = new BitArray(320000);

// Traverse array elements
for (int i=0; i<arr.length; i++)
{
// Index in bit array
int num  = arr[i] - 1;

// If num is already present in bit array
if (ba.get(num))
System.out.print(num +" ");

// Else insert num
else
ba.set(num);
}
}

// Driver code
public static void main(String[] args) throws
java.lang.Exception
{
int[] arr = {1, 5, 1, 10, 12, 10};
checkDuplicates(arr);
}
}
```

Output:

```1 10
```

This article is contributed by Mr. Somesh Awasthi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
4.3 Average Difficulty : 4.3/5.0
Based on 13 vote(s)