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I am given an input, "N", i have to find the number of list of length N, which starts with 1, such that the next number to be added is at most 1 more than the max number added till now. For Example,

N = 3, possible lists => (111, 112, 121, 122, 123), [113, or 131 is not possible as while adding '3' to the list, the maximum number present in the list would be '1', thus we can add only 1 or 2].

N = 4, the list 1213 is possible as while adding 3, the maximum number in the list is '2', thus 3 can be added.

Problem is to count the number of such lists possible for a given input "N".

My code is :- 

public static void Main(string[] args) { var noOfTestCases = Convert.ToInt32(Console.ReadLine()); var listOfOutput = new List<long>(); for (int i = 0; i < noOfTestCases; i++) { var requiredSize = Convert.ToInt64(Console.ReadLine()); long result; const long listCount = 1; const long listMaxTillNow = 1; if (requiredSize < 3) result = requiredSize; else { SeqCount.Add(requiredSize, 0); AddElementToList(requiredSize, listCount, listMaxTillNow); result = SeqCount[requiredSize]; } listOfOutput.Add(result); } foreach (var i in listOfOutput) { Console.WriteLine(i); } } private static Dictionary<long, long> SeqCount = new Dictionary<long, long>(); private static void AddElementToList(long requiredSize, long listCount, long listMaxTillNow) { if (listCount == requiredSize) { SeqCount[requiredSize] = SeqCount[requiredSize] + 1; return; } var listMaxTillNowNew = listMaxTillNow + 1; for(var i = listMaxTillNowNew; i > 0; i--) { AddElementToList(requiredSize, listCount + 1, i == listMaxTillNowNew ? listMaxTillNowNew : listMaxTillNow); } return; }

Which is the brute force method. I wish to know what might be the best algorithm for the problem? PS : I only wish to know the number of such lists, so i am sure creating all the list won't be required. (The way i am doing in the code) I am not at all good in algorithms, so please excuse for the long question.

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    Is this a homework? Is the text you posted exactly the question? I ask because I don't properly understand the question, and if it is homework, you might post the exact question. Commented Apr 7, 2012 at 19:47
  • No its not a homework, its just a puzzle a friend asked me, If you have some doubts about the question, please ask, may be i can clarify. Commented Apr 7, 2012 at 19:50
  • Okay - is 111 a list of three values, or is it just one number? Commented Apr 7, 2012 at 19:51
  • That is a list, all those are examples of valid list for the input length of 3. Thus, for input = 3, there are 5 different lists that are possible. Commented Apr 7, 2012 at 19:53
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    You want to compute the nth Bell number. oeis.org/A000110 Commented Apr 7, 2012 at 19:59

2 Answers 2

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This problem is a classic example of a dynamic programming problem:

If you define a function dp(k, m) to be the number of lists of length k for which the maximum number is m, then you have a recurrence relation:

dp(1, 1) = 1 dp(1, m) = 0, for m > 1 dp(k, m) = dp(k-1, m) * m + dp(k-1, m-1)

Indeed, there is only one list of length 1 and its maximum element is 1. When you are building a list of length k with max element m, you can take any of the (k-1)-lists with max = m and append 1 or 2 or .... or m. Or you can take a (k-1)-list with max element m-1 and append m. If you take a (k-1)-list with max element less than m-1 then by your rule you can't get a max of m by appending just one element.

You can compute dp(k,m) for all k = 1,...,N and m = 1,...,N+1 using dynamic programming in O(N^2) and then the answer to your question would be

dp(N,1) + dp(N,2) + ... + dp(N,N+1)

Thus the algorithm is O(N^2).

See below for the implementation of dp calculation in C#:

 int[] arr = new int[N + 2]; for (int m = 1; m < N + 2; m++) arr[m] = 0; arr[1] = 1; int[] newArr = new int[N + 2]; int[] tmp; for (int k = 1; k < N; k++) { for (int m = 1; m < N + 2; m++) newArr[m] = arr[m] * m + arr[m - 1]; tmp = arr; arr = newArr; newArr = tmp; } int answer = 0;strong text for (int m = 1; m < N + 2; m++) answer += arr[m]; Console.WriteLine("The answer for " + N + " is " + answer);
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2 Comments

This surely gives the correct answer, thanks for the code, but is it possible to do this with a lesser complexity. O(n) may be.
I don't think you can go below O(n^2). The problem is that in order to determine what are the possible values for k-th element you need to know the maximum value within the first k-1. And since there are k possibilities you need to sum k different numbers. To reach N you need to do this N times - so quadratic is the best you can hope for.
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Well, I got interrupted by a fire this afternoon (really!) but FWIW, here's my contribution:

 /* * Counts the number of possible integer list on langth N, with the * property that no integer in a list(starting with one) may be more * than one greater than the greatest integer preceeding it in the list. * * I am calling this "Semi-Factorial" since it is somewhat similar to * the factorial function and its constituent integer combinations. */ public int SemiFactorial(int N) { int sumCounts = 0; // get a list of the counts of all valid lists of length N, //whose maximum integer is listCounts[maxInt]. List<int> listCounts = SemiFactorialCounts(N); for (int maxInt = 1; maxInt <= N; maxInt++) { // Get the number of lists, of length N-1 whose maximum integer //is (maxInt): int maxIntCnt = listCounts[maxInt]; // just sum them up sumCounts += maxIntCnt; } return sumCounts; } // Returns a list of the counts of all valid lists of length N, and //whose maximum integer is [i], where [i] is also its index in this //returned list. (0 is not used). public List<int> SemiFactorialCounts(int N) { List<int> cnts; if (N == 0) { // no valid lists, cnts = new List<int>(); // (zero isn't used) cnts.Add(0); } else if (N == 1) { // the only valid list is {1}, cnts = new List<int>(); // (zero isn't used) cnts.Add(0); //so that's one list of length 1 cnts.Add(1); } else { // start with the maxInt counts of lists whose length is N-1: cnts = SemiFactorialCounts(N - 1); // add an entry for (N) cnts.Add(0); // (reverse order because we overwrite the list using values // from the next lower index.) for (int K = N; K > 0; K--) { // The number of lists of length N and maxInt K { SF(N,K) } // Equals K times # of lists one shorter, but same maxInt, // Plus, the number of lists one shorter with maxInt-1. cnts[K] = K * cnts[K] + cnts[K - 1]; } } return cnts; }

pretty similar to the others. Though I wouldn't call this "classic dynamic programming" so much as just "classic recursion".

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