Find the missing quantity

Introduction to algorithms Chapter 4 questions 4-2:

An array a [1. N] contains all integers ranging from 0 to N, but one of them is no longer in the array. In this case, we cannot use a single operation to access a complete integer in A. the only operation we can use is to take the J-bits of a [I ", the time taken by this operation is a constant. An algorithm is designed to locate the missing integer in the O (n) time. Any complete integer other than a can be accessed by a single operation.

Idea: the topic of Chapter 4 is "recursive". This problem may be caused by the Division and control method, and the scale of the problem may be reduced by some means. The division symbol "/" used below is the computer integer division operator.

(1) For the percentile bits of All integers in A, if A is ordered, 0 and 1 are alternating. The total number of 0 and the total number of 1 are at most 1 different. 0 ~ can be determined based on the parity of N ~ The number of random digits 0 of N and the number of 1. Then, after performing a round of statistics on the percentile bit of a [1. N], we can see whether the percentile bit of the missing number is 0 or 1.

A) if n is an even number, then 0 ~ The number of random digits 0 of N is n/2 + 1, and the number of 1 is n/2. After a round of statistics, if the number of missing integers is 0, we only need to continue to investigate these array items with the number of missing integers: Ignore the number of missing integers, the problem is to find the number missing from 0 to n/2 in this subarray. If we find that the number is 1, similarly, as long as you continue to investigate the array items with the first digit: Ignore the second digit, the problem is converted to finding the number of missing values from 0 to n/2-1 in this subarray.

The recursion formula is T (n) = T (n/2) + N.

PS: If n is an even number, then 0 ~ N all the even numbers are divided by 2, and the obtained sequence is 0 ~ N/2; 0 ~ All the odd numbers in N are divided by 2, and the obtained sequence is 0 ~ N/2-1.

If n is an odd number,

Then 0 ~ N all the even numbers are divided by 2, and the obtained sequence is 0 ~ N/2; 0 ~ All the odd numbers in N are divided by 2, and the obtained sequence is 0 ~ N/2.

 

B) N is an odd number, similar to.

 

 

(2) consider the highest bit. Suppose K = [log2

N]-an integer that contains k + 1 digits. So 0 ~ The number of the highest bits in N is 2 K.

(0 ~

2 K

-1)

, The maximum number of 1 is the n-2k

.

After a round of statistics on the highest digit, we can conclude that the missing number is between 0 and ~ 2 K

-Within 1, or 2 K

~ N, if the former ignores the highest bit, the problem is converted to finding

0 ~ 2 K

-1;

If it is the latter, ignore the highest bit. The problem is to find 0 ~ in the sub-array ~ N-2k

.

 

 

Another question:

How can I find the number missing in a [1... n.

(1) to 0 ~ N is summed, and the sum of the data in a is subtracted.

(2) Use an exclusive or operation. 0 ^ 1 ^... ^ n ^ A [1] ^... ^ A [n] is the number of missing values.

(3) After the end of a fast sorting round, if Element B of the central axis is either B + 1 or B, the problem is converted to a [B + 2... search for B + 1 ~ in N ~ Number of missing values in N; if the latter, the problem is converted to a [1 ~ 0 ~ In the b-1 ~ Number of defects in the B-1.