Hi guys i just want to know if it is possible to find the rank of a string. Is it possible?

Question

Given a word all the letters of it are permuted in all possible ways and thus formed words are arranged as in a dictionary. Then the position of the given word is simply called rank of the word.
Procedure to find the rank of a word
Step 1: Write the word and above it,put the alphabetical ordering.
Step 2:Write below,the factorial values starting with 0!(=1) from right to left.
Step 3:Write the number of numbers less than(in top row) the number towards right of it.
Step 4:Multiply the numbers shown by and add.
Step 5:Add "1" to get the rank of the word.
Example: The rank of the word "MASTER" is
Step 1: 3 1 5 6 2 4
M A S T E R
Step 2: 120 24 6 2 1 1
Step 3: 2 0 2 2 0 0
Step 4: 240+0+12+4+0+0=256
Step 5: 256+1=257

Therefore 257 is the rank of the word "MASTER"

Answer 1

Marking so I can get back to this. This can be a fun exercise, and yes, it is perfectly possible. Characters are linked to their ASCII values.

Answer 2

@Hatsy Rei Thanks.

Answer 3

Shamefully I only managed to get to step 1, failed to understand the numbers outputted on successive steps :|

Good idea though, bookmarking the thread :)

Answer 4

@RR2001, thanks! found one to read on quora just now, gonna read about it :)

Answer 5

@lpang Its quite simple. Just google permutation and combination, how to find the rank of a word. That one was just an example

Answer 6

@Gordie
Now I get it! Thanks!

Ill simply use this to get the count:
for(int i=0;i<str.size();i++)
{
ctr[i] = count_if(a.begin()+1,a.end(),
[](char c){return c<a[i];})
}

Answer 7

There is one more method to find the rank but before that you need to find the factorial of a number. EX, factorial of 5 is , 5! = 5 x 4 x 3x 2 x 1 =120,
Now the question would be find the dictionary rank of the word LIFT
ANS
First of all arrange in alphabetical order, F I L T
Now no of words beginning with F (I,L,T) - 3! = 6
no of words beginning with I (F,L,T) - 3! = 6
since L is the first letter of the word whose rank is what we are concerned L is locked and the our second unknown letter will be F
L x x x
No of words beginning with F = 2! = 2
I is the second letter of the word whose rank is what we are concerned so I is also locked
L I x x
if proceed in the original order the subsequent letter is F, so lock F
L I F x
Next letter finally turns out to be T
L I F T.
so add total = 6 + 6+ 2 = 14
as lift is also a word we need to consider that too
14 + 1 = 15.
Rank of lift is 15.

While codes on this topic are welcome, from my side i am working on this method, as this method needs loops to work out. Interested coders can try this method and the other method and also write a code for finding the rank of a word having repeating letters(not distinct). The concept varies for this code and i recommend you to watch a video since it is hard to explain via words

Answer 8

What do we have to do in step 3? value[pos]-value[pos+1]? Can someone explain me step 3 if possible?

Answer 9

here is my try
https://code.sololearn.com/cIJE3G1EAZ9q/?ref=app

Answer 10

yes