fairy hunter - 1 year ago 91

Java Question

I have an infinite String. The length of this string is infinite in our imagination and cannot be limited.Suppose that we have a sequence String like this:

"123456789..."

The

"...7891011121314..."

In this section, I want to explain about the requirement. The requirement is to find the index of first occurrence of input String (called

n="3"

First Index of n = 2

n="910"

First Index of n = 8

I have coded the algorithm to find the index of the String

Snippet of my Code:

`while(!num.contains(s)){`

num +=start.toString();

start = start.add(BigInteger.ONE);

}

This is my full code: Full Code of My Code

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Answer Source

Here's a general description of how to solve this problem. Translating it to Java can still be challenging.

Your input String is basically an infinite sequence of all the natural numbers 1 2 3 4 5 6 8 9 10 11 ....

I believe the point of the exercise is to identify the first sub-sequence of natural numbers of the input String that the substring `n`

belongs to, and then calculate its index without actually constructing the large "infinite" String.

For that purpose, you have to try to split the substring `n`

into an incrementing sequence of numbers having as few digits as possible.

First you have to check is the substring `n`

creates a sequence of single digits numbers. That is the case, for example, if n == 345678 (note that n may contain both single digit and double digit numbers, for example `n == 345678910`

, which you should also be able to identify).

If you fail in that step, you should look for a sequence of double digit numbers. That is the case, for example, if `n == 33343536`

. Now, this could get trickier, since `n == 2333435363`

is also a sequence of two digit numbers, but the leading and trailing numbers of the sequence (32 and 37) are truncated.

If you fail again, you look for a sequence of 3 digit numbers.

If you don't find any sequence, you treat the entire substring `n`

as a single number in the big String.

Now, suppose `n`

is `199319941995`

, and you discovered in the previous step that the first number in the sequence is `1993`

. The remaining work is to calculate the index of the number `1993`

in the input String. You know that the single digit numbers take 1*9 indices. The two digit numbers take 2*90 indices. The three digit numbers take 3*900 indices. The 3 digit numbers between 1000 and 1993 take 4*993 indices. Therefore the index of 1993 is 1*9+2*90+3*900+4*993, and that's the first index of the substring `199319941995`

.

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