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Amazon Interview Question for Software Engineer/Developer about Algorithms, Bit Magic
(16 posts)-
An array A[1...n] contains all the integers from 0 to n except for one number which is missing. In this problem, we cannot access an entire integer in A with a single operation. The elements of A are represented in binary, and the only operation we can use to access them is “fetch the jth bit of A[i]”, which takes constant time. Write code to find the missing integer. Can you do it in O(n) time?
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divide the numbers based on bits
first 0th bit -- all odd numbers (having 0th bit 1) and even numbers(having 0th bit 0)
we are left with 2 parts
find umber from which part is missing
then in that subpart check for 1st bit again by dividing number into two partscomplexity - n+n/2+n/4+n/8 so on approx equal to 2n = O(n)
with an example
lets take number upto
5 with 3 missing1 - 001
2- 010
// 3-011(missing)
4-100
5-101
6-110
divide into two parts1 -001 2 -010
5 -101 4 -100
6-110
one from left side is missing
now apply algo using 1st bit
001
101
as both bits are zero 1st bit of missing number should be 1
so we have 11 known now for 2nd bit again apply algo
as there is no number with 1st bit 1 other than this number in this series so 2nd bit shall be zero as number r starting frm 1 and number is 011 i.e 3 -
XOR all j bits for A[1..n] together.
The result is the missing number. -
@shandon your complexity will be O(jn) not O(n) and j has to be the bits in highest value of a[i] here
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@raj..This is true. Could you explain code-wise what you mean by dividing into two parts broseph?
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Nevermind I get it.
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@rajcool explin
what it is
"find umber from which part is missing
then in that subpart check for 1st bit again by dividing number into two parts"what u r doing here find umber from which part is missing
then in that subpart check for 1st bit again by dividing number into two parts -
@raja -- i am dividing the number into two parts one part in which particular bit is 1 and other in which that bit is 0
the part that contain fewer elements contains missing element[you ofcourse have to take into consideration whether number of elements are odd or even at start]
we first check for 0th bit divide given array into two parts
then on part in which missing element is we apply same algo for 1st bit again divide into parts and then repeat process for 2nd bit and so on -
what if the nos are odd?
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i mean if the number of elements is odd ?
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An array A[1 n] contains all the integers from 0 to n except one. It would be easy
to determine the missing integer in O(n) time by using an auxiliary array B[0 n]
to record which numbers appear in A. In this problem, however, we cannot
access an entire integer in A with a single operation. The elements of A are
represented in binary, and the only operation we can use to access them is
"fetch the jth bit of A[i]," which takes constant time. Show that if we use only
this operation, we can still determine the missing integer in O(n) time.
1.Parse all the element of the array looking for it zeroth bit.
2.Calculate number of zeros and ones encountered and its index in the array.
3.if number of zeros are greater than ones
4.Parse all the element whose ending bit is 1. and look for it's second last bit.
5.Repeat the above steps for number of bits. -
@hatella:
According your algo i hope time complexity is not O(n) it is O(jn) isn't it ? -
didnt get it :( @rajacool ...btw will your algo work if arr[] has element in unsorted order???
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PsychoCoder
Member
Posted 3 months ago #
Here j is constant for 32 bit numbers!!
So basically it still O(n) !!
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Dedicated Programmer
Member
Posted 1 month ago #
@rajcools , i am a little bit confused.
What to do if the number of elements is odd, i mean it has not been mentioned in the question that n will be even.
So,in this case,dividing the array into two halves will always result into either
(number of 0's > number of 1's) or
(number of 1's > number of 0's).
So, what to do in this case. -
Dedicated Programmer
Member
Posted 1 month ago #
@Shandon , here elements can be accessed bit by bit,
So,before Xoring all the elements,first we need to access them one by one and store them somewhere.
Don't you think that it will result into extra space complexity.
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