Find Duplicates of array using bit array

08 Jul, 2022
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++
Java
Python3
C#
Javascript
// 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 integersclass 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 codeint 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 program to print all Duplicates in arrayimport java.util.*;import java.lang.*;import java.io.*; // A class to represent array of bits using// array of integersclass 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 program to print all Duplicates in array # A class to represent array of bits using# array of integersclass 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 Duplicatesdef 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 Codeif __name__ == "__main__":    arr = [1, 5, 1, 10, 12, 10]    checkDuplicates(arr) # This code is contributed by# sanjeev2552
// C# program to print all Duplicates in array  // A class to represent array of bits using// array of integersusing 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 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 Duplicatesfunction 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 codevar a = [ 1, 5, 1, 10, 12, 10 ];checkDuplicates(a); // This code is contributed by phasing17.
1 10 
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Rajput-Ji
sanjeev2552
raghavgehlot682
phasing17
hardikkoriintern
Arrays
Bit Magic
Arrays
Bit Magic
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