# Find Duplicates of array using bit array

• Difficulty Level : Hard
• Last Updated : 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```

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 ``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);``    ``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

## 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 `

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