Find Duplicates of array using bit array
Last Updated :
23 Feb, 2023
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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++ 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
// C++ program to print all Duplicates in arrayusing 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 array
import java.util.*;
import java.lang.*;
import java.io.*;
// A class to represent array of bits using
// array of integers
public 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 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# 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 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.
let index = (pos >> 5);
// Now find bit number in arr[index]
let bitNo = (pos & 0x1F);
// Find value of given bit number in
// arr[index]
let 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 Kirti Agarwal(kirtiagarwal23121999)
Output
1 10
Time Complexity: O(N)
Space Complexity: O(1)
