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 |
// 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) |
1 10
Time Complexity: O(N)
Space Complexity: O(1)