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Find Duplicates of array using bit array in Python
Suppose we have an array of n different numbers; n can be 32,000 at max. The array may have duplicate entries and we do not know what is the value of n. Now if we have only 4-Kilobytes of memory, how would display all duplicates in the array?
So, if the input is like [2, 6, 2, 11, 13, 11], then the output will be [2,11] as 2 and 11 appear more than once in given array.
To solve this, we will follow these steps −
Create one byte-array type data structure bit_arr, it has following methods
Define constructor This will take n
arr := an array of size (n / 2^5) + 1, fill with 0
Define a function get_val() . This will take pos
index := pos / 2^5
bitNo := pos AND 31
return true when (arr[index] AND (2^bitNo)) is not same as 0
Define a function set_val() . This will take pos
index := pos / 2^5
bitNo := pos AND 31
arr[index] := arr[index] OR (2^bitNo)
From the main method, do the following −
arr := bit_arr(320000)
for i in range 0 to size of arr, do
num := arr[i]
if arr.get_val(num) is non-zero, then
display num
otherwise,
set_val(num) of arr
Example
Let us see the following implementation to get better understanding −
class bit_arr: def __init__(self, n): self.arr = [0] * ((n >> 5) + 1) def get_val(self, pos): self.index = pos >> 5 self.bitNo = pos & 31 return (self.arr[self.index] & (1 << self.bitNo)) != 0 def set_val(self, pos): self.index = pos >> 5 self.bitNo = pos & 31 self.arr[self.index] |= (1 << self.bitNo) def is_duplicate(arr): arr = bit_arr(320000) for i in range(len(arr)): num = arr[i] if arr.get_val(num): print(num, end = " ") else: arr.set_val(num) arr = [2, 6, 2, 11, 13, 11] is_duplicate(arr)
Input
[2, 6, 2, 11, 13, 11]
Output
2 11
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