Alternative solutions for additional questions for an interview

I haven't written a blog for a while. During a stroll today, I saw a blog titled "A company held an annual" Developer "gathering in South China in April 7-recording a company's written test ". There are Interview Questions recalled by the author. One of the questions attracted the attention of the author. The questions are as follows:

Question: An array a [n] is known to construct an array B [N]. The construction rules are as follows: B [I] = A [0] * A [1] * A [2]... A [n]/A [I];
Requirements:
1. Division is not allowed;
2. the time complexity is O (n), and the space complexity is S (0 );
3. You cannot use other variables except for the variables used for traversal;

 

It seems simple, and it's a waste of time to think about it.

The first thought is the method that the original author thought of. The Code is as follows (vb2008 is used ):

Public shared function cacub1 (byval A () as double ()
Dim B (A. Length-1) as double
Dim I as integer
B (0) = 1
For I = 0 to A. Length-1
B (0) * = a (I)
Next
For I = A. Length-1 to 0 step-1
B (I) = B (0)/A (I)
Next
Return B
End Function

This method is concise with no additional code. The only thing that does not meet the requirement is that division is used.

The Code is as follows:

Public shared function cacub2 (byval A () as double ()
Dim B (A. Length-1) as double
Dim I as integer, J as integer
For I = 0 to A. Length-1
B (I) = 1
For J = 0 to A. Length-1
If I <> J then B (I) * = a (j)
Next
Next

Return B
End Function

Although the calculation amount is increased, division is not used. However, the time complexity of the algorithm is O (n * n), which does not meet the question requirements. This method is also rigid and is not recommended by the author.

 

After thinking for a long time, there is always an imbalance between division and time complexity.

Suddenly, a thought flashed through. Division? How about turning a bend? What is the conversion to subtraction?

Use formula S/a = 10lgs-lga

So this question becomes

S = a (0) * A (1) * A (2 )...... * A (n)

B (I) = 10lgs-lga (I)

The Code is as follows:

Public shared function cacub3 (byval A () as double ()
Dim B (A. Length-1) as double
Dim I as integer
B (0) = 1
For I = 0 to A. Length-1
B (0) * = a (I)
Next
For I = A. Length-1 to 0 step-1
B (I) = 10 ^ (math. log10 (B (0)-math. log10 (A (I )))
Next
Return B
End Function

 

Have you met the question requirements? Yes, no division is used, and the time complexity is O (n ). The efficiency is a little lower.

But is efficiency really low? Not necessarily. vs optimizes the log10 function. Although it is a little lower, it is negligible.

This problem is solved by the logarithm formula of the high school stage. If anyone has a better algorithm, I hope you will be enlightened. Everyone learns from each other and improves together.