![]() String word = br.readLine(). Int cases = Integer.parseInt(br.readLine()) Putting it all together: public static void main(String args) throws IOException (Last one is just general good practice.) Finally, always declare variables as close as possible to where they are actually used. Use a loop to multiply the factorials together. The letters construct is trickery that makes the count for A appear at index 0, B at index 1 etc. Just loop through the characters and count them. Also, string replacement to check character counts is overly complicated. You can keep the letter counts in an array of length 26. ![]() I also want to know about other ways of solving this program. How can I make this program shorter and more efficient? ("Data set %d: %d\n",iteration,result) īut I think it's very lengthy and not efficient. Result = factorial(lengthOfWord) / (factorial(a)*factorial(b)*factorial(c)*factorial(d)*factorial(e)*factorial(f)*factorial(g)*factorial(h)*factorial(i)*factorial(j)*factorial(k)*factorial(l)*factorial(m)*factorial(n)*factorial(o)*factorial(p)*factorial(q)*factorial(r)*factorial(s)*factorial(t)*factorial(u)*factorial(v)*factorial(w)*factorial(x)*factorial(y)*factorial(z)) Z = lengthOfWord - word.replace("Z", "").length() ![]() Y = lengthOfWord - word.replace("Y", "").length() X = lengthOfWord - word.replace("X", "").length() W = lengthOfWord - word.replace("W", "").length() V = lengthOfWord - word.replace("V", "").length() U = lengthOfWord - word.replace("U", "").length() T = lengthOfWord - word.replace("T", "").length() S = lengthOfWord - word.replace("S", "").length() R = lengthOfWord - word.replace("R", "").length() Q = lengthOfWord - word.replace("Q", "").length() P = lengthOfWord - word.replace("P", "").length() O = lengthOfWord - word.replace("O", "").length() N = lengthOfWord - word.replace("N", "").length() M = lengthOfWord - word.replace("M", "").length() L = lengthOfWord - word.replace("L", "").length() K = lengthOfWord - word.replace("K", "").length() J = lengthOfWord - word.replace("J", "").length() I = lengthOfWord - word.replace("I", "").length() H = lengthOfWord - word.replace("H", "").length() G = lengthOfWord - word.replace("G", "").length() Int a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z īufferedReader br = new BufferedReader(new InputStreamReader(System.in)) įor(int iteration =1 iteration <= cases iteration++)Ī = lengthOfWord - word.replace("A", "").length() ī = lengthOfWord - word.replace("B", "").length() Ĭ = lengthOfWord - word.replace("C", "").length() ĭ = lengthOfWord - word.replace("D", "").length() Į = lengthOfWord - word.replace("E", "").length() į = lengthOfWord - word.replace("F", "").length() ![]() Example 8.12: In how many ways can be letters of the word 'GLOOMY'be. Public static void main(String args) throws IOException (a) 2n+1 (b) 22n+1 (c) 2n1 The problem on Permutation and Combination are broadly. Here's my code: import java.io.BufferedReader My process is to find the factorial of the length of the word,then dividing it by the multiplication of the factorials of the occurrences of different alphabets present in the word.(using recursion). So that computing it would take a lot of effort.I am writing a program in java that will count the number of ways a word can be permuted/rearranged.įor example the word "HAPPY" can be rearranged in 60 ways. The answer then is 4!!!! The number is quite big (how big?). two followed by two factorials.Īnd finally, 0 = 0 followed by zero factorials - a result of doing nothing. three followed by three factorials.Ģ = 2! = (2!)!, i.e. (n - 1)! ways to count an n-element set.For each of these, by definition, the remaining (n-1) elements can be counted in (n-1)! ways. There are n ways to select the first element. Let's try mimicking this for a set of n elements. Look at the six permutations of a 3-element set. The remaining element automatically goes to the fourth place. For example, if you have just been invited to the Oscars and you have only 2 tickets for friends and family to bring with you, and you have 10 people to choose. There remain only three candidates for the second position and, after this was selected, only two candidates for the third position. A permutation is a way to select a part of a collection, or a set of things in which the order matters and it is exactly these cases in which our permutation calculator can help you. What's 4!? There are 4 ways to select the first element. I placed the answer to the question at the bottom of this page. Guess the next number in the following sequence However, the result of this activity is nothing or, in math parlance, 0. There is just one way to do nothing so that 0! = 1. ) Since there is nothing to count the question is In how many ways can one do nothing? A mathematical answer to this is just one: 0! = 1. Many interesting questions in probability theory require us to calculate the number of ways You can arrange a set of objects. In other words, the set V can be brought into a 1-1 correspondence with the set. A set V consists of n elements if its elements can be counted 1, 2., n.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |