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 duplicate entries and you do not know what N is. With only 4 Kilobytes of memory available, how would print all duplicate 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

Asked In: Amazon 

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 the implementation of the idea. 

Implementation:

C++

// C++ program to print all Duplicates in array
#include <bits/stdc++.h>
using namespace std;

// A class to represent an array of bits using
// array of integers
class BitArray
{
    int *arr;

    public:
    BitArray() {}

    // Constructor
    BitArray(int n)
    {
        // Divide 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
    bool 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
    void checkDuplicates(int arr[], int n)
    {
        // create a bit with 32000 bits
        BitArray ba = BitArray(320000);

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

            // If num is already present in bit array
            if (ba.get(num))
                cout << num << " ";

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

// Driver code
int main()
{
    int arr[] = {1, 5, 1, 10, 12, 10};
    int n = sizeof(arr) / sizeof(arr[0]);
    BitArray obj = BitArray();
    obj.checkDuplicates(arr, n);
    return 0;
}

// This code is contributed by
// sanjeev2552

Java

// 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)
    {
        // Divide 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);
    }
}

Python3

# Python3 program to print all Duplicates in array

# A class to represent array of bits using
# array of integers
class BitArray:

    # Constructor
    def __init__(self, n):

        # Divide by 32. To store n bits, we need
        # n/32 + 1 integers (Assuming int is stored
        # using 32 bits)
        self.arr = [0] * ((n >> 5) + 1)

    # Get value of a bit at given position
    def get(self, pos):

        # Divide by 32 to find position of
        # integer.
        self.index = pos >> 5

        # Now find bit number in arr[index]
        self.bitNo = pos & 0x1F

        # Find value of given bit number in
        # arr[index]
        return (self.arr[self.index] &
                   (1 << self.bitNo)) != 0

    # Sets a bit at given position
    def set(self, pos):

        # Find index of bit position
        self.index = pos >> 5

        # Set bit number in arr[index]
        self.bitNo = pos & 0x1F
        self.arr[self.index] |= (1 << self.bitNo)

# Main function to print all Duplicates
def checkDuplicates(arr):

    # create a bit with 32000 bits
    ba = BitArray(320000)

    # Traverse array elements
    for i in range(len(arr)):

        # Index in bit array
        num = arr[i]

        # If num is already present in bit array
        if ba.get(num):
            print(num, end = " ")

        # Else insert num
        else:
            ba.set(num)

# Driver Code
if __name__ == "__main__":
    arr = [1, 5, 1, 10, 12, 10]
    checkDuplicates(arr)

# This code is contributed by
# sanjeev2552

C#

// C# program to print all Duplicates in array

// A class to represent array of bits using
// array of integers
using System;

class BitArray
{
    int[] arr;

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

    // Get value of a bit at given position
    bool 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];

            // If num is already present in bit array
            if (ba.get(num))
                Console.Write(num + " ");

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

    // Driver code
    public static void Main()
    {
        int[] arr = {1, 5, 1, 10, 12, 10};
        checkDuplicates(arr);
    }
}

// This code is contributed by Rajput-Ji

Javascript

// JavaScript program to print all Duplicates in array

// A class to represent array of bits using
// array of integers

class BitArray {
    // Constructor
    constructor(n)
    {
        // Divide by 32. To store n bits, we need
        // n/32 + 1 integers (Assuming int is stored
        // using 32 bits)
        this.arr = new Array((n >> 5) + 1);
    }
    // Get value of a bit at given position
    get(pos)
    {
        // Divide by 32 to find position of
        // integer.
        var index = (pos >> 5);

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

        // Find value of given bit number in
        // arr[index]
        var arrCopy = this.arr return (arrCopy[index]
                                       & (1 << bitNo))
                      != 0;
    }
    // Sets a bit at given position
    set(pos)
    {
        // Find index of bit position
        var index = (pos >> 5);

        // Set bit number in arr[index]
        var bitNo = (pos & 0x1F);
        var arr1 = this.arr;
        arr1[index] = arr1[index] | (1 << bitNo);
        this.arr = arr1;
    }
}

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

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

        // If num is already present in bit array
        if (ba.get(num))
            console.log(num);

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

// Driver code
var a = [ 1, 5, 1, 10, 12, 10 ];
checkDuplicates(a);

// This code is contributed by phasing17.

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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.