Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
621 views
in Technique[技术] by (71.8m points)

algorithm - Find the index of a given permutation in the sorted list of the permutations of a given string

We're given a string and a permutation of the string.

For example, an input string sandeep and a permutation psdenae.

Find the position of the given permutation in the sorted list of the permutations of the original string.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

The total number of permutation of a given string of length n would be n! (if all characters are different), thus it would not be possible to explore all the combinations.

This question is actually like the mathematics P & C question

Find the rank of the word "stack" when arranged in dictionary order.

Given the input string as NILSU Take a word which we have to find the rank. Take "SUNIL" for example.

Now arrange the letter of "SUNIL" in alphabetical order.

It will be. "I L N S U".

Now take the first letter. Its "I". Now check, is the letter "I" the first letter of "SUNIL"? No. The number of words that can be formed starting with I will be 4!, so we know that there will be 4! words before "SUNIL".

I = 4! = 24

Now go for the second letter. Its "L". Now check once again if this letter we want in first position? No. So the number of words can be formed starting with "L" will be 4!.

L = 4! = 24

Now go for "N". Is this we want? No. Write down the number of words can be formed starting with "N", once again 4!

N = 4! = 24

Now go for "S". Is this what we want? Yes. Now remove the letter from the alphabetically ordered word. It will now be "I L N U"

Write S and check the word once again in the list. Is we want SI? No. So the number of words can be formed starting with SI will be 3!

[S]:I-> 3! = 6

Go for L. is we want SL? No. So it will be 3!.

[S]:L-> 3! = 6

Go for N. is we want SN? No.

[S]:N-> 3! = 6

Go for SU. Is this we want? Yes. Cut the letter U from the list and then it will be "I L N". Now try I. is we want SUI? No. So the number of words can be formed which starts from SUI will be 2!

[SU]:I-> 2! = 2 Now go for L. Do we want "SUL". No. so the number of words starting with SUL will be 2!.

[SU]:L-> 2! = 2

Now go for N. Is we want SUN? Yes, now remove that letter. and this will be "I L". Do we want "SUNI"? Yes. Remove that letter. The only letter left is "L".

Now go for L. Do we want SUNIL? Yes. SUNIL were the first options, so we have 1!. [SUN][I][L] = 1! = 1

Now add the whole numbers we get. The sum will be.

24 + 24 + 24 + 6 + 6 + 6 + 2 + 2 + 1 = 95.

So the word SUNIL will be at 95th position if we count the words that can be created using the letters of SUNIL arranged in dictionary order.

Thus through this method you could solve this problem quite easily.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

1.4m articles

1.4m replys

5 comments

56.8k users

...