I read this problem once… but i noticed that was above my level, and i focused in Problem B. Reading now… i notice that i did a good choice, because is a very hard problem.

Only 1366 users did it the small. And only 31 of them did the large one.

If some of you have some idea of this problem, i appreciate some of info, or some approach!

Thank you very much

Maybe i did a totally wrong approach, i don´t know. I needed to change my code at minute 23 when an announcement appeared, and i mean that i add wrong code there.

Only 2323 users did the small, and 456 of them did the large one.

Here my C# code:

using System;
using System.Collections.Generic;
using System.IO;

namespace GoogleCodeJam
{
    class Program
    {
        static void Main(string[] args)
        {
            StreamReader sr = new StreamReader(@"C:\B-small-attempt10.in");
            StreamWriter sw = new StreamWriter(@"C:\5.txt");
            string numberTest = sr.ReadLine();
            string[] temp = new string[int.Parse(numberTest) * 2];
            for (int i = 0; i < temp.Length; i++)
                temp[i] = sr.ReadLine();
            int cases = 1;
            int maxEnergy = 0;
            for (int i = 0; i < temp.Length; i += 2)
            {
                string[] temp1 = temp[i].Split(' ');
                string[] temp2 = temp[i + 1].Split(' ');
                maxEnergy = (int.Parse(temp1[1]) > int.Parse(temp1[0])) ? int.Parse(temp1[1]) : int.Parse(temp1[0]);
                int energy = int.Parse(temp1[0]);
                for (int a = 0; a < temp2.Length; a++)
                {
                    if (int.Parse(temp2[a]) == 1)
                    {
                        energy -= maxEnergy;
                        if (a + 1 != temp2.Length)
                            energy += int.Parse(temp1[1]);
                    }
                    else
                    {
                        energy += maxEnergy * int.Parse(temp2[a]);
                    }

                }
                sw.WriteLine("Case #" + cases + ": " + energy.ToString());
                cases++;
            }
            sw.Close();

        }
    }
}

If you have some clean approach, or you know why my code is wrong, i will be very grateful for your explanation.

Thank you very much for all your help

I tried to do this problem in Round 1A, but i read and read again, and I think that I couldn´t even understand well…  I will read the analisys when availabe, but until then, if some of you want to explain me, or give me his/her approach, i will be very grateful.

Only 5856 users did the small, and 1806 of them the large one.

Thank you very much

Here i decided to in a new string variable, store the new letters, then with a for loop, check every leeter in the input, and then, if is a vowel or ‘y’, take only the first one. I do this with a bool variable, when i get the first i put the bool to true, then i only take a new wovel if the bool is false, and i put the bool to false when i add a new consonant. Then when the for loop ends, return the length of the new string as an int.

Here the problem statement:

Elly recently learned that vowels (the letters ‘a’, ‘e’, ‘i’, ‘o’, ‘u’ and ‘y’) are no longer cool on the internet. That’s why now she wants to remove some of the vowels in the nickname she uses. Elly wants her new nickname to contain no consecutive vowels, but she still wants to keep as many characters as possible.

You are given the current nickname Elly uses in the string nickname. Return its new length after it gets shortened.

Here my code:

public class EllysNewNickname
    {
        public int getLength(string nickname)
        {
            string r = "";
            bool x = false;
            for (int i = 0; i < nickname.Length; i++)
            {
                if (nickname[i] != 'a' && nickname[i] != 'e' && nickname[i] != 'i' && nickname[i] != 'o' && nickname[i] != 'u' && nickname[i] != 'y')
                {
                    r += nickname[i].ToString();
                    x = false;
                }
                else
                {
                    if (x == false)
                    {
                        r += nickname[i].ToString();
                        x = true;
                    }
                }
            }
            return r.Length;
        }
    }

If you have another approach, please share with me in a comment!

Thank you very much

I would like to thank all of you.. Everyone of you!

THANK YOU!

Now.. i wil continue working for all of you… I am really happy for this!

• vCenter 5.1 Update1
• ESXi 5.1 Update1
• SRM 5.1.1 and vSphere Replication 5.1.1
• vSphere Data Protection (VDP) Version 5.1.10
• vCloud Director 5.1.2

What’s New in vCenter 5.1 Update1?

• vCenter can now be installed on Windows Server 2012
• vSphere Essentials and Essentials Plus are no longer limited to 192Gb vRAM
• Support of SQL Server 2012 and SQL Server 2008R2SP2 database for vCenter
• Support for Gust Customization

What’s New in VDP 5.10?

• Integration with vCenter alarms and alerts notification system
• Ability to clone backup jobs
• New filters for Restore tab
• Added restore rehearsal
• VDR to VDP Migration Utility NOT available in VDP advanced

Whats’s New in vCloud Director 5.1.2?

• Rights for creating, reverting, and removing snapshots
• Allocation pool organization virtual datacenters can be elastic or non-elastic
• Support for Microsoft SQL Server 2012
• Support for Red Hat Enterprise Linux 6.3
• Guest Customization support on Windows Server 2012

Read this post to know about the feature and fixes included in SRM5.1.1 and vSphere Replication 5.1.1

Make sure to read all the release notes before upgrading to update1

Happy Upgrading

Tonight’s April Dirty Thirty beer is…

Sixpoint Bengali Tiger

Sixpoint Bengali Tiger Ale

Critical Information:

• 6.4% ABV
• 62 IBUs
• 13 SRM

My Experience:

- POUR

Poured from can into pint glass. It has an orange color and is mostly transparent. It has a large, puffy white head that clings to the glass.

- SMELL

It has an aroma of fresh hops and is citrus-y.

- TASTE

It’s bitter, then hoppy, then bitter again. It’s straw-like (is that something??) and has a hint of grapefruit.

- FOOD PAIRING

Gouda cheese or Peruvian chicken. Fried yuca anyone?

- OVERALL THOUGHTS

Disappointed. I’ve been seeing the Beer Advocate guys and some bloggers running around touting Sixpoint. I still really want to try 3Beans but didn’t enjoy tonight’s Bengali Tiger that much.

Score: 3/10

Thank you for being a part of the April Dirty Thirty project. Keep the tweets and comments coming, you’re doing outstanding! Remember to use #AprilDirty30. I’d love to hear your thoughts on this beer, what you had tonight, what you’ve tried recently, anything.

One new beer a night for a month. Review it. Share it. Expand your beer knowledge.

Some of the fixed issues, for example, are things like:

• All sorts of timeout problems ranging from multiple operation timeouts to reprotect timeouts to HBA rescan timeouts
• Custom vCenter https ports now work better with vSphere Replication
• Pairing SRM servers using custom certificates and VCVA now works
• Re-protect using vSphere Replication is more resilient

Get the release notes  for VMware vCenter Site Recovery Manager 5.1.1 from here.

Get the release notes  for VMware vSphere Replication 5.1.1 from here.

Here i decided to convert the input to seconds, then with a for loop, take each second beginning with i = 1, then do (i * 100) / totalSecs, and check if contains a “.”, (if is a double), if it is, then continue, else, add +1 to the final result. When the for loop ends, return the result as an int.

Here the problem statement:

You are writing firmware for an exercise machine. Each second, a routine in your firmware is called which decides whether it should display the percentage of the workout completed. The display does not have any ability to show decimal points, so the routine should only display a percentage if the second it is called results in a whole percentage of the total workout.

Given a string time representing how long the workout lasts, in the format “hours:minutes:seconds”, return the number of times a percentage will be displayed by the routine. The machine should never display 0% or 100%.

Here my code:

public class ExerciseMachine
    {
        public int getPercentages(string time)
        {
            int res = 0;
            double temp = 0.0;
            string[] secs = time.Split(':');
            double sec = (double.Parse(secs[0]) * 3600.0) + (double.Parse(secs[1]) * 60.0) + double.Parse(secs[2]);
            for (int i = 1; i < sec; i++)
            {
                temp = ((double)i * 100.0) / sec;
                if (!(temp.ToString().Contains(".")))
                    res++;
            }
            return res;
        }
    }

If you have another approach, please share with me in a comment!

Thank you very much

The problem

Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = { S1, S2, .. , Sm} valued coins, print all thw ays in which we can make the change.

For example, for N = 4 and S = {1,2,3}, there are four solutions: {1,1,1,1},{1,1,2},{2,2},{1,3}. For N = 10 and S = {2, 5, 3, 6}, there are five solutions: {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} and {5,5}.

The code

#include<iostream.h>
#include<conio.h>
#include<vector.h>

void print_possibilities(int* coins,vector<int> &str,int k,int sum,int step)
{
     if(sum==k)
     {
               for(int i=0;i<str.size();i++)
               cout<<str[i]<<"+";
               cout<<"="<<k<<endl;          
     } 
     else
     {
         for(int i=step;i<3;i++)
         {
                 if(sum+coins[i]<=k)
                 {
                          str.push_back(coins[i]);
                          print_possibilities(coins,str,k,sum+coins[i],i);     
                          str.pop_back();                        
                 }        
                 else 
                 {
                      break;
                 }
         }    
     }   
}

main()
{
     vector<int> str;
     int coins[]={1,2,5};
     print_possibilities(coins,str,7,0,0);
     getch();      
}

If you have some questions, let them in a comment and Karanpreet will answer asap.

Thank you

The problem

#include<iostream.h>
#include<conio.h>
#include<vector.h>

void func(int n,int len,vector<int> &str,int *a)
{
        if(n==len)
        {
                  int i=0;
                  for(int i=0;i<str.size();i++){
                  cout<<str[i]<<" ";}
                  cout<<endl;          
        }  
        else
        {
            func(n+1,len,str,a);
            str.push_back(a[n]);
            func(n+1,len,str,a);
            str.pop_back();    
        }
}

main()
{
      vector<int> str;
      int a[]={1,2,3,4};
      func(0,4,str,a);
      getch();      
}

Here first i decide to store in a int[] array calles total, the number os times each number repeat in the input. And put 6 and 9 together in the 6. Then, reduce the 6 doing, if total[6] % 2 == 0, then total[6] = total[6] / 2, else total[6] = (total[6] / 2) + 1.Then only get the higher num in the array and return it as an int.

Here the problem statement:

You are going to stick the number of your room on the door. The shop near your house suggests wonderful sets of plastic digits. Each set contains exactly ten digits – one of each digit between 0 and 9, inclusive. Return the number of sets required to write your room number. Note that 6 can be used as 9 and vice versa. 

Here my code:

public class RoomNumber
    {
        public int numberOfSets(int roomNumber)
        {
            int[] tot = new int[10];
            for (int i = 0; i < roomNumber.ToString().Length; i++)
            {
                if (roomNumber.ToString()[i] == '9')
                    tot[6]++;
                else
                    tot[int.Parse(roomNumber.ToString()[i].ToString())]++;
            }
            if (tot[6] % 2 == 0)
                tot[6] = tot[6] / 2;
            else
                tot[6] = (tot[6] / 2) + 1;
            int res = -1;
            for (int i = 0; i < tot.Length; i++)
            {
                if (tot[i] > res)
                    res = tot[i];
            }
            return res;
        }
    }

If you have another approach, please share with me in a comment!

Thank you very much

Here i decided first convert the two strings toLower(). Then with regular expressions, and with a foreach loop in the search string, go checking each letter, then count the number of occurrences in each word, and then do Math.Pow(n1 – n2, 2), and add to the result. Then return the result.

Here the problem statement:

The distance between two strings with respect to a letter is defined as (n₁ – n₂)², where n₁ and n₂ are the number of occurrences (both lowercase and uppercase) of that letter in the first and second strings, respectively. The distance between two strings with respect to a third string is the sum of the distances between the two strings with respect to each letter in the third string.

You will be given three strings a, b and letterSet. All the letters in letterSet will be distinct. Return the distance between a and b with respect to letterSet. 

Here my code:

public class DistanceBetweenStrings
    {
        public int getDistance(string a, string b, string letterSet)
        {
            int res = 0;
            a = a.ToLower();
            b = b.ToLower();
            foreach (char c in letterSet)
            {
                Regex r = new Regex(c.ToString());
                res += int.Parse(Math.Pow((double)r.Matches(a).Count - (double)r.Matches(b).Count, 2).ToString());
            }
            return res;
        }
    }

If you have another approach, please share with me in a comment!

Thank you very much

Here i decided to use regular expression to get all the index where the searched word appear… then beginning with i = 1, check if i >= i – 1, if it is, then add +1 to the result. We need to take care if there are 0 coincidences, or only 1, in these cases only return 0 or 1 in each case.

Here the problem statement:

You have been tasked with writing a function that will scan through a given document, and determine how many times a given word or phrase appears in that document. However, it is important that your function does not count overlapping occurrences. For instance, if the document were “abababa”, and the search keyword was “ababa”, you could find the keyword starting at index 0, or at index 2, but not both, since they would overlap.

You must concatenate the elements of the given string[] doc into a single string. Then, return the maximum number of non-overlapping occurrences of the string search.

To find a maximal set of non-overlapping occurrences, perform the following procedure. Starting from the left, find the first occurrence of the search string. Then, continuing with the character immediately following the search string, continue looking for the next occurrence. Repeat until no new occurrences can be found. By continuing immediately following each found occurrence, we guarantee that we will not count overlaps. 

Here the code:

public class DocumentSearch
    {
        public int nonIntersecting(string[] doc, string search)
        {
            int res = 0;
            Regex r = new Regex(search);
            string total = "";
            for (int i = 0; i < doc.Length; i++)
                total += doc[i];
            List<int> list = new List<int>();
            foreach (Match m in r.Matches(total))
                list.Add(m.Index);
            if (list.Count == 0)
                return 0;
            if (list.Count == 1)
                res++;
            else
            {
                res++;
                for (int a = 1; a < list.Count; a++)
                {
                    if (list[a] >= list[a - 1] + search.Length)
                        res++;
                }
            }
            return res;
        }
    }

If you have another approach, please share with me in a comment!

Thank you very much

The code

#include<iostream.h>
#include<conio.h>
#include<vector.h>

void func(int left,int right,vector<char> str)
{
     //if right becomes zero you have to just print the string bcos this is the last step of recursion, kind of base case
     if(right==0)
     {
                 for(int i=0;i<str.size();i++)
                 {
                         cout<<str[i];        
                 }            
                 cout<<endl;
                 return;
     }
     //if there are still left parenthesis which can be used , use them!
     if(left>0)
     {
               str.push_back('(');
               func(left-1,right,str);
               str.pop_back();
     }
     //this means more of left has been used and right can be used now
     if(left<right)
     {
                   str.push_back(')');
                   func(left,right-1,str);    
                   str.pop_back();          
     }
}

main()
{
      vector<char> str;
     // we are generating all the valis combinatiuons of 3 pairs of parenthesis, so 3 is passed as left        and right to the function
      func(3,3,str);
      getch();      
}

If you have some questions, let them in a comment and Karanpreet will answer asap.

Thank you

The problem

#include<iostream.h>
#include<conio.h>

int ctr=0;
struct node
{
       int n;
       node* next;      
};

node* head=NULL;

void insert(int data)
{
     ctr++;
     node* p=(node*)malloc(sizeof(node));
     p->n=data;
     if(head==NULL)
     {
                   head=p;
                   head->next=NULL;
                   }    
     else
     {
          node* temp=head;
          head=p;
          head->next=temp;
     }
}

int get_min(node* elem)
{
    if(elem==NULL)
    return 32767;
    int min_from_rec=get_min(elem->next);
    if(min_from_rec>elem->n)
    return elem->n;
    else
    return min_from_rec;
}

main()
{
      insert(2);
      insert(4);
      insert(-1);
      insert(3);
      insert(5);
      insert(-8);
      insert(6);
      cout<<get_min(head)<<endl<<ctr;
      getch();      
}

The problem

#include<iostream.h>
#include<conio.h>

int func(int* a,int len,int &start,int &end)
{
    int tempsum=0,maxsum=0;
    for(int i=0;i<len;i++)
    {
             tempsum=tempsum+a[i];
             if(tempsum>0)
             {
                          if(tempsum>maxsum)
                          {
                              maxsum=tempsum;   
                              end=i; 
                              }         
             }       
             else
             {
                 tempsum=0;  
                 start=i+1;  
             }
    }    
    return maxsum;
}

main()
{
      int a[]={-2,3,6,-8,7,6,3,-8};
      int start, end;
      cout<<"The maximum sum is "<<func(a,8,start,end);
      cout<<" starting at "<<start<<" ending at "<<end;
      getch();      
}

Here i decided to check each letter, f is ‘N’, then if contain ‘S’, remove the ‘S’ and the ‘N’, and do this with all the letters, then when the for loop ends, convert to array, sort, and convert back to a string to return.

Here the problem statement:

Billy is going to his grandmother’s house. To help him do that, his mother has written down a detailed list of instructions for him to follow. Each instruction is a character ‘N’, ‘S’, ‘W’ or ‘E’, telling him to go exactly 1 block to the north, south, west or east, respectively. Billy’s city consists of an infinitely large grid of streets, where each street extends infinitely to both sides, and the space between 2 adjacent streets going in the same direction is always 1 block. Billy’s house and his grandmother’s house are both located at street corners in this city.

Billy knows that his mother does not always choose the shortest path. Therefore, he wants to make a new list of instructions that will also lead him to his grandmother’s house, but uses the minimum possible number of instructions.

You will be given inst, a string with the original list made by Billy’s mom. Return the new list Billy wants. If there are several solutions, return the one that comes first alphabetically. 

Here my code:

public class OptimalList
    {
        public string optimize(string inst)
        {
            for (int i = 0; i < inst.Length; i++)
            {
                if (inst[i] == 'N')
                {
                    if (inst.Contains("S"))
                    {
                        inst = inst.Remove(i, 1);
                        inst = inst.Remove(inst.IndexOf("S"), 1);
                        i = -1;
                    }
                }
                else if (inst[i] == 'S')
                {
                    if (inst.Contains("N"))
                    {
                        inst = inst.Remove(i, 1);
                        inst = inst.Remove(inst.IndexOf("N"), 1);
                        i = -1;
                    }
                }
                else if (inst[i] == 'E')
                {
                    if (inst.Contains("W"))
                    {
                        inst = inst.Remove(i, 1);
                        inst = inst.Remove(inst.IndexOf("W"), 1);
                        i = -1;
                    }
                }
                else if (inst[i] == 'W')
                {
                    if (inst.Contains("E"))
                    {
                        inst = inst.Remove(i, 1);
                        inst = inst.Remove(inst.IndexOf("E"), 1);
                        i = -1;
                    }
                }
            }
            string[] temp = new string[inst.Length];
            for (int i = 0; i < inst.Length; i++)
                temp[i] = inst[i].ToString();
            Array.Sort(temp);
            string res = "";
            for (int i = 0; i < temp.Length; i++)
                res += temp[i];
            return res;
        }
    }

If you have another approach, please share with me in a comment!

Thank you very much

Here only follow the steps in the problem statement. You have all the clues there.

Here the problem statement:

You wish to buy a new car and would like to get the cheapest option available. However, you recognize that the initial purchase cost is only a part of what you will need to spend on the car and you want to factor tax and fuel costs into your decision. You will be given a string[] cars giving the specifications of various models of car and an int annualDistance containing the distance that you will drive each year. For the car that you choose, the purchase cost only has to be paid once, the tax cost specific to the model has to be paid once per year and the fuel cost should be calculated as FUEL PRICE * DISTANCE DRIVEN / FUEL EFFICIENCY, where the fuel efficiency is specific to the car. You should calculate which car model in cars has the lowest cost on aggregate after years amount of time and return a double containing this cost.

Each element of cars will be formatted “<PURCHASE COST> <TAX> <FUEL EFFICIENCY>” (quotes for clarity). 

Here my code:

public class CarBuyer
    {
        public double lowestCost(string[] cars, int fuelPrice, int annualDistance, int years)
        {
            double result = 10000000000.0;
            for (int i = 0; i < cars.Length; i++)
            {
                double res = 0.0;
                string[] temp = cars[i].Split(' ');
                res += double.Parse(temp[0]);
                res += double.Parse(temp[1]) * (double)years;
                res += ((double)fuelPrice * ((double)annualDistance * (double)years)) / double.Parse(temp[2]);
                if (res < result)
                    result = res;
            }
            return result;
        }
    }

If you have another approach, please share with me in a comment!

Thank you very much

Here i decided to use a for loop, fron 0, to n.Length – 1, then check if mine[i] == ‘<’ && mine[i + 1] == ‘>’ then remove the diamond from the mine, add +1 to the final result, and put 1 to -1 to begin the loop again. When the for loop ends, return the result as an int.

Here the problem statement:

You are a diamond hunter looking for diamonds in a peculiar mine. This mine is a string of ‘<’ and ‘>’ characters, and each diamond is a substring of the form “<>”. Each time you find a diamond, you remove it and the residual mine is updated by removing the 2 characters from the string.

For example, if you have a mine like “><<><>>><”, you can start by removing the first appearance of “<>” to get “><<>>><”, then remove the only remaining diamond to get “><>><”. Note that this produces a new diamond which you can remove to get “>><”. Since there are no diamonds left, your expedition is done.

Given a string mine, return the number of diamonds that can be found. Note that the order in which you remove simultaneous appearances of diamonds is irrelevant; any order will lead to the same result. 

Here my code:

public class DiamondHunt
    {
        public int countDiamonds(string mine)
        {
            int res = 0;
            if(mine.Length == 1)
                return 0;
            for (int i = 0; i < mine.Length - 1; i++)
            {
                if (mine[i] == '<' && mine[i + 1] == '>')
                {
                    mine = mine.Remove(i, 2);
                    res++;
                    i = -1;
                }
            }
            return res;
        }
    }

If you have another approach, please share with me in a comment!

Thank you very much

The problem

Given a string, find its rank among all its permutations sorted lexicographically. For example, rank of “abc” is 1, rank of “acb” is 2, and rank of “cba” is 6.

#include<iostream.h>
#include<conio.h>

struct node
{
       char data;
       node* next;       
};

node* head=NULL;
node* tail=NULL;

void insert(char a)
{
     node* p=new node;
     p->data=a; p->next=NULL;
     if(head==NULL)
     {head=p;tail=p;}
     else
     {
         tail->next=p;
         tail=tail->next;    
     }     
}

void remove(char c)
{
     node* p=head,*q=head;
     while(p!=NULL)
     {
                   if(p->data==c)
                   break;
                   q=p;
                   p=p->next;              
     }
     if(p==head)
     {head=head->next;}
     else
     q->next=p->next;
}

void iterate()
{
   cout<<endl;
     node* p=head;
     while(p!=NULL)
     {
                   cout<<p->data;
                   p=p->next;              
     }          
}

void initialize()
{
     insert('a');     
     insert('b');     
     insert('c');     
     insert('d');     
}

int distance(char a)
{
    node* p=head; int j=0;
    while(p!=NULL && p->data!=a)
    {j++;p=p->next;}   
    remove(a); 
    return j;
}

int fact(int n)
{
    if(n==1 || n==0)
    return 1;
    else
    return n*fact(n-1);    
}

int rank(char *a,int len)
{
        int i=0,res=0,step;
        while(i<4)
        {
                 step=distance(a[i]);
                 res=res+step*fact(len-i);
                 i++;                 
        }
        return res+1;
}

main()
{
      char *a="dbca";
      initialize();
      cout<<"The lexicographic rank is "<<rank(a,3);
      getch();    
}

If you have some questions, please let them in a comment, and Karan will answer asap.

Thank you

#include<iostream.h>
#include<conio.h>

void swap(char *a,char *b)
{
     char temp;
     temp=*a;
     *a=*b;
     *b=temp;     
}

void perm(char* a,int i,int n)
{
          if(i==n-1)
          {
                  cout<<a<<endl;
                  return;
          }
          for(int j=i;j<n;j++)
          {
                  swap(&a[j],&a[i]);
                  perm(a,i+1,n);
                  swap(&a[j],&a[i]);
                  }
}

main()
{
      //if we have the character array already sorted then the permutations are automatically in lexicographic order
      char a[]="abc";
      perm(a,0,3);
      getch();      
}

If you have some questions, leave them in a comment and Karanpreet will answer asap!

Thank you