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.
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 |
Output:
1 10
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