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Removing even/odd numbers in problem 5.7 of cracking the coding interview


Forum Post 0 Answers Removing even/odd numbers in problem 5.7 of cracking the coding interview

Following is problem 5.7 (under Bit Manipulation) in the 5th edition of Cracking the coding interview:

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 opera-tion. 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?

The algorithm applied is this:

1. Start with LSB.
2. Count occurrence of 1's vs 0's.
If count(1) < count(0) it means the missing number has a 1 as it's LSB, else it has a 0.
3. Remove all numbers with LSB not matching result found in step 3.
4. Repeat steps 1 to 4, and progressively checking the next LSB in each iteration.

Can someone explain the logic behind step 3? It basically removes all odd/even numbers from the current list (depending on the bit found for the missing number) and uses the modified list in the next iteration. Why do we do this?

 - shahsunny712 May 14, 2015 | Flag |


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Step 3 removes non-candidate numbers. If a missing number ith bit is 0, it cannot be any numbers whose ith bit is 1. By removing number whose ith bit is 1, it reduces the candidate number by half.
Example: Suppose the missing number V is 000. if 0th bit of V is 0, V cannot be any numbers whose 0th bit is 1. Thus 001 and 011 are removed.
- - -
001
010
011

- ibn Humboldt July 21, 2015 | Flag Reply
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testing

- ss July 21, 2015 | Flag Reply
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msg

- po July 21, 2015 | Flag Reply
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In my opinion, the question is very poorly worded. It says the numbers can only be accessed one bit at a time in constant time. But then allows copying the numbers into two different arrays in O(n) time (each number is copied in constant time) instead of O (n log n) time (where each number needs to have all it's bits copied in O(log n) time).

- Jackfruit September 30, 2018 | Flag Reply




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