Finding the ranking of a word (permutations) with duplicate letters
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Finding the ranking of a word (permutations) with duplicate letters
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 the rank of the Given string in list of all possible permutations with Duplicates) did not work for me.
So I tried this (which is a compilation of code I've plagiarized and my attempt to deal with repetitions). The non-repeating cases work fine. BOOKKEEPER generates 83863, not the desired 10743.
(The factorial function and letter counter array 'repeats' are working correctly. I didn't post to save space.)
while (pointer != length) { if (sortedWordChars[pointer] != wordArray[pointer]) { // Swap the current character with the one after that char temp = sortedWordChars[pointer]; sortedWordChars[pointer] = sortedWordChars[next]; sortedWordChars[next] = temp; next++; //For each position check how many characters left have duplicates, //and use the logic that if you need to permute n things and if 'a' things //are similar the number of permutations is n!/a! int ct = repeats[(sortedWordChars[pointer]-64)]; // Increment the rank if (ct>1) { //repeats? System.out.println("repeating " + (sortedWordChars[pointer]-64)); //In case of repetition of any character use: (n-1)!/(times)! //e.g. if there is 1 character which is repeating twice, //x* (n-1)!/2! int dividend = getFactorialIter(length - pointer - 1); int divisor = getFactorialIter(ct); int quo = dividend/divisor; rank += quo; } else { rank += getFactorialIter(length - pointer - 1); } } else { pointer++; next = pointer + 1; } } Answer by David Eisenstat for Finding the ranking of a word (permutations) with duplicate letters
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 charCounts = new java.util.HashMap(); 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 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 by John for Finding the ranking of a word (permutations) with duplicate letters
@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 Counter(string word) { char[] wordArray = word.ToCharArray(); int len = word.Length; SortedDictionary count = new SortedDictionary(); 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 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 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 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 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 by Peiti Peter Li for Finding the ranking of a word (permutations) with duplicate letters
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 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 by Shubham Nanda for Finding the ranking of a word (permutations) with duplicate letters
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 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 by Kiran for Finding the ranking of a word (permutations) with duplicate letters
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)
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