/*

  The function f is defined on any finite sequence of integer numbers.

  It returns total number of 3-length segments of the sequence containing “0 1 2” 

  (e.g. number of indexes k where a[k]=0, a[k+1]=1, and a[k+2]=2).

  Find inductive extension of f. 

*/                  

using System;

using System.Diagnostics;

public class Program

{

    public static void Main()

    {

        // Just for testing - modify to test other cases

        int [] a = { 0, 1, 2, 0, 0, 1, 1, 2, 0, 1, 2, 2, 0};

        Console.WriteLine(f(a, 3));

    }

    // We'll solve more common problem and we'll find number of segments containing

    // "0 1 .. s-1"

    static int f(int [] a, int s)

    {

        Trace.Assert(s > 0);

        // Inductive extension:

        int n   = 0;    // number of completed sequences 0 .. s-1

        int len = 0;    // length of last incompleted sequence 0 .. s-1, len < s

        foreach(int nextElem in a)  // we can read array same way as we'd read user input

        {

            if(nextElem == len)

            {   // nextElem extends last segment

                len++;

            }

            else

            {   // nextElem does not extend last segment

                if(nextElem == 0)

                    len = 1;

                else

                    len = 0;

            }

            // nextElem processed with respect to last segment

            if(len == s)

            {   // last segment is completed

                n++;

                len = 0;

            }

        }

        return n;

    }

}