Finding the ranking of a word (permutations) with duplicate letters

Question

I\'m posting this although much has already been posted about this question. I didn\'t want to post as an answer since it\'s not working. The answer to this post (Finding th

Answer 1

Note: this answer is for 1-based rankings, as specified implicitly by example. Here's some Python that works at least for the two examples provided. The key fact is that suffixperms * ctr[y] // ctr[x] is the number of permutations whose first letter is y of the length-(i + 1) suffix of perm.

from collections import Counter

def rankperm(perm):
    rank = 1
    suffixperms = 1
    ctr = Counter()
    for i in range(len(perm)):
        x = perm[((len(perm) - 1) - i)]
        ctr[x] += 1
        for y in ctr:
            if (y < x):
                rank += ((suffixperms * ctr[y]) // ctr[x])
        suffixperms = ((suffixperms * (i + 1)) // ctr[x])
    return rank
print(rankperm('QUESTION'))
print(rankperm('BOOKKEEPER'))

Java version:

public static long rankPerm(String perm) {
    long rank = 1;
    long suffixPermCount = 1;
    java.util.Map<Character, Integer> charCounts =
        new java.util.HashMap<Character, Integer>();
    for (int i = perm.length() - 1; i > -1; i--) {
        char x = perm.charAt(i);
        int xCount = charCounts.containsKey(x) ? charCounts.get(x) + 1 : 1;
        charCounts.put(x, xCount);
        for (java.util.Map.Entry<Character, Integer> e : charCounts.entrySet()) {
            if (e.getKey() < x) {
                rank += suffixPermCount * e.getValue() / xCount;
            }
        }
        suffixPermCount *= perm.length() - i;
        suffixPermCount /= xCount;
    }
    return rank;
}

Unranking permutations:

from collections import Counter

def unrankperm(letters, rank):
    ctr = Counter()
    permcount = 1
    for i in range(len(letters)):
        x = letters[i]
        ctr[x] += 1
        permcount = (permcount * (i + 1)) // ctr[x]
    # ctr is the histogram of letters
    # permcount is the number of distinct perms of letters
    perm = []
    for i in range(len(letters)):
        for x in sorted(ctr.keys()):
            # suffixcount is the number of distinct perms that begin with x
            suffixcount = permcount * ctr[x] // (len(letters) - i)
            if rank <= suffixcount:
                perm.append(x)
                permcount = suffixcount
                ctr[x] -= 1
                if ctr[x] == 0:
                    del ctr[x]
                break
            rank -= suffixcount
    return ''.join(perm)

Answer 2

@Dvaid Einstat, this was really helpful. It took me a WHILE to figure out what you were doing as I am still learning my first language(C#). I translated it into C# and figured that I'd give that solution as well since this listing helped me so much!

Thanks!

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Text.RegularExpressions;

namespace CsharpVersion
{
    class Program
    {
        //Takes in the word and checks to make sure that the word
        //is between 1 and 25 charaters inclusive and only
        //letters are used
        static string readWord(string prompt, int high)
        {
            Regex rgx = new Regex("^[a-zA-Z]+$");
            string word;
            string result;
            do
            {
                Console.WriteLine(prompt);
                word = Console.ReadLine();
            } while (word == "" | word.Length > high | rgx.IsMatch(word) == false);
            result = word.ToUpper();
            return result;
        }

        //Creates a sorted dictionary containing distinct letters 
        //initialized with 0 frequency
        static SortedDictionary<char,int> Counter(string word)
        {
            char[] wordArray = word.ToCharArray();
            int len = word.Length;
            SortedDictionary<char,int> count = new SortedDictionary<char,int>();
           foreach(char c in word)
           {
               if(count.ContainsKey(c))
               {
               }
               else
               {
                   count.Add(c, 0);
               }

           }
           return count;
        }

        //Creates a factorial function
        static int Factorial(int n)
        {
            if (n <= 1)
            {
                return 1;
            }
            else
            {
                return n * Factorial(n - 1);
            }
        }
        //Ranks the word input if there are no repeated charaters 
        //in the word
        static Int64 rankWord(char[] wordArray)
        {
            int n = wordArray.Length; 
            Int64 rank = 1; 
            //loops through the array of letters
            for (int i = 0; i < n-1; i++) 
            { 
                int x=0; 
            //loops all letters after i and compares them for factorial calculation
                for (int j = i+1; j<n ; j++) 
                { 
                    if (wordArray[i] > wordArray[j]) 
                    {
                        x++;
                    }
                }
                rank = rank + x * (Factorial(n - i - 1)); 
             }
            return rank;
        }

        //Ranks the word input if there are repeated charaters
        //in the word
        static Int64 rankPerm(String word) 
        {
        Int64 rank = 1;
        Int64 suffixPermCount = 1;
        SortedDictionary<char, int> counter = Counter(word);
        for (int i = word.Length - 1; i > -1; i--) 
        {
            char x = Convert.ToChar(word.Substring(i,1));
            int xCount;
            if(counter[x] != 0) 
            {
                xCount = counter[x] + 1; 
            }
            else
            {
               xCount = 1;
            }
            counter[x] = xCount;
            foreach (KeyValuePair<char,int> e in counter)
            {
                if (e.Key < x)
                {
                    rank += suffixPermCount * e.Value / xCount;
                }
            }

            suffixPermCount *= word.Length - i;
            suffixPermCount /= xCount;
        }
        return rank;
        }




        static void Main(string[] args)
        {
           Console.WriteLine("Type Exit to end the program.");
           string prompt = "Please enter a word using only letters:";
           const int MAX_VALUE = 25;
           Int64 rank = new Int64();
           string theWord;
           do
           {
               theWord = readWord(prompt, MAX_VALUE);
               char[] wordLetters = theWord.ToCharArray();
               Array.Sort(wordLetters);
               bool duplicate = false;
               for(int i = 0; i< theWord.Length - 1; i++)
               {
                 if(wordLetters[i] < wordLetters[i+1])
                 {
                     duplicate = true;
                 }
               }
               if(duplicate)
               {
               SortedDictionary<char, int> counter = Counter(theWord);
               rank = rankPerm(theWord);
               Console.WriteLine("\n" + theWord + " = " + rank);
               }
               else
               {
               char[] letters = theWord.ToCharArray();
               rank = rankWord(letters);
               Console.WriteLine("\n" + theWord + " = " + rank);
               }
           } while (theWord != "EXIT");

            Console.WriteLine("\nPress enter to escape..");
            Console.Read();
        }
    }
}

Answer 3

Java version of unrank for a String:

public static String unrankperm(String letters, int rank) {
    Map<Character, Integer> charCounts = new java.util.HashMap<>();
    int permcount = 1;
    for(int i = 0; i < letters.length(); i++) {
        char x = letters.charAt(i);
        int xCount = charCounts.containsKey(x) ? charCounts.get(x) + 1 : 1;
        charCounts.put(x, xCount);

        permcount = (permcount * (i + 1)) / xCount;
    }
    // charCounts is the histogram of letters
    // permcount is the number of distinct perms of letters
    StringBuilder perm = new StringBuilder();

    for(int i = 0; i < letters.length(); i++) {
        List<Character> sorted = new ArrayList<>(charCounts.keySet());
        Collections.sort(sorted);

        for(Character x : sorted) {
            // suffixcount is the number of distinct perms that begin with x
            Integer frequency = charCounts.get(x);
            int suffixcount = permcount * frequency / (letters.length() - i); 

            if (rank <= suffixcount) {
                perm.append(x);

                permcount = suffixcount;

                if(frequency == 1) {
                    charCounts.remove(x);
                } else {
                    charCounts.put(x, frequency - 1);
                }
                break;
            }
            rank -= suffixcount;
        }
    }
    return perm.toString();
}

See also n-th-permutation-algorithm-for-use-in-brute-force-bin-packaging-parallelization.

Answer 4

I would say David post (the accepted answer) is super cool. However, I would like to improve it further for speed. The inner loop is trying to find inverse order pairs, and for each such inverse order, it tries to contribute to the increment of rank. If we use an ordered map structure (binary search tree or BST) in that place, we can simply do an inorder traversal from the first node (left-bottom) until it reaches the current character in the BST, rather than traversal for the whole map(BST). In C++, std::map is a perfect one for BST implementation. The following code reduces the necessary iterations in loop and removes the if check.

long long rankofword(string s)
{
    long long rank = 1;
    long long suffixPermCount = 1;
    map<char, int> m;
    int size = s.size();
    for (int i = size - 1; i > -1; i--)
    {
        char x = s[i];
        m[x]++;
        for (auto it = m.begin(); it != m.find(x); it++)
                rank += suffixPermCount * it->second / m[x];

        suffixPermCount *= (size - i);
        suffixPermCount /= m[x];
    }
    return rank;
}

Answer 5

If there are k distinct characters, the i^th character repeated n_i times, then the total number of permutations is given by

            (n_1 + n_2 + ..+ n_k)!
------------------------------------------------ 
              n_1! n_2! ... n_k!

which is the multinomial coefficient.
Now we can use this to compute the rank of a given permutation as follows:

Consider the first character(leftmost). say it was the r^th one in the sorted order of characters.

Now if you replace the first character by any of the 1,2,3,..,(r-1)^th character and consider all possible permutations, each of these permutations will precede the given permutation. The total number can be computed using the above formula.

Once you compute the number for the first character, fix the first character, and repeat the same with the second character and so on.

Here's the C++ implementation to your question

#include<iostream>

using namespace std;

int fact(int f) {
    if (f == 0) return 1;
    if (f <= 2) return f;
    return (f * fact(f - 1));
}
int solve(string s,int n) {
    int ans = 1;
    int arr[26] = {0};
    int len = n - 1;
    for (int i = 0; i < n; i++) {
        s[i] = toupper(s[i]);
        arr[s[i] - 'A']++;
    }
    for(int i = 0; i < n; i++) {
        int temp = 0;
        int x = 1;
        char c = s[i];
        for(int j = 0; j < c - 'A'; j++) temp += arr[j];
        for (int j = 0; j < 26; j++) x = (x * fact(arr[j]));
        arr[c - 'A']--;
        ans = ans + (temp * ((fact(len)) / x));
        len--;
    }
    return ans;
}
int main() {
    int i,n;
    string s;
    cin>>s;
    n=s.size();
    cout << solve(s,n);
    return 0;
}

Answer 6

If we use mathematics, the complexity will come down and will be able to find rank quicker. This will be particularly helpful for large strings. (more details can be found here)

Suggest to programmatically define the approach shown here (screenshot attached below) given below)