How do I write a program in C to find the rank of a word (with repetition)?

#include long fact(int n) { long f=1; while (n > 0) { f *= n; n--; } return f; } int find_rank(char s[], int len) { long rank = 1; int i, j, k, l, fr[26], f; //fr is frequency char sorted[100], cur, t, pre; for (i = 0; i < len -1 ; ++i) { cur = s[i]; strcpy(sorted, s + i); l = len - i; memset(fr, 0, sizeof(fr)); fr[sorted[0] - 'a']++; for (j = 1; j < l; ++j) { t=sorted[j]; fr[t-'a']++; for (k = j-1; k>=0 && sorted[k] > t; --k) sorted[k+1] = sorted[k]; sorted[k+1] = t; } pre = 1; for (k = 0; sorted[k] != cur ;++k) { if (sorted[k] == pre) continue; pre = sorted[k]; f = fact(l - 1); fr[pre - 'a']--; for (j = 0; j < 26; ++j) if (fr[j]) f /= fact(fr[j]); rank += f; fr[pre - 'a']++; } } return rank; } int main() { char s[100]; scanf("%s", s); printf("%d", find_rank(s, strlen(s))); return 0; } read less

lets see how to find the rank of word if the letter in a word having repetition. Before preceding to the tutorial some of the basics we should know. Like how many total words can be formed using letters of INDIA. In letter “INDIA” there are 2 I and no other letter is repeating. So the no. of words that can form using letter of INDIA will be 5! / 2! = 60. Now we may proceed to the next point. Algorithm 1. Take that word which we have to find the rank. Now the letter of that word is repeating. I take “BOMBAY” 2. Now arrange the letter of “BOMBAY” in alphabetical order. Now they are “ABBMOY“. 3. Pickup first letter from “ABBMOY” that’s “A“. Now compare “A” from “BOMBAY”. Does the letter “A” in the first we want? NO. Now how many words can be made if we extract “A” from “ABBMOY“. A = 5! / 2! = 60 Now proceed to the next. 4. Pickup second letter from “ABBMOY” that’s “B“. Now compare “B” from “BOMBAY“. Does the letter “B” in the first we want? Yes. Now fix the letter “B“. [B] Then go to once again on the first letter of “ABBMOY”. Does “A” we want? NO. Then write down how many words can be made with starting [B]A [B]A = 4! = 24 5. Repeat the process until we get the letter which we want after B in the word “BOMBAY“. [B]B = 4! = 24 [B]M = 4! = 24 [B]O. “O” we want. So fix this and get back to the starting. 6. Does we want now [BO]A? NO. So write down once again. [BO]A = 3! = 6 [BO]B = 3! = 6 [BO]M. “M” we want now. So fix this too. And get back to the starting. 7. Does we want [BOM]A? No. So write down once again. [BOM]A = 2! = 2 [BOM]B. “B” we want. So fix this too. And get back to starting. We see “A” also we want now. So fix this too. And the last letter is “Y” We also want this. So we reach to the end. Now we can write. [BOMBAY] = 1. 8. Now add the entire sum which we earn. That’s 60 + 24 +24 + 24 + 6 + 6 + 2 + 1 = 147. That’s the answer. Now for Word “INDIA” 1. [ADIIN] 2. A = 4! / 2! = 12 2. D = 4! / 2! = 12 3. [I]A = 3! = 6 4. [I]D = 3! = 6 5. [I]I = 3! = 6 6. [IN]A = 2! =2 7. [IND]IA = 1. Now add all. 12 + 12 + 6 + 6 + 6 + 2 + 1 = 45. So that’s the answer. Congratulation now you are able to find RANK of any word if its letters having repetition or not. In the case of repetition there is no shortcut. read less