Problem Statement
We have a rectangular castle. The first floor is protected by some number of guards. We want to have at least one guard in each row and in each column. You are given a String[] castle. The j-th character of the i-th element of castle is either ‘.’ (free cell) or uppercase ‘X’ (guard). Return the smallest number of additional guards we have to place in the castle to achieve our goal.

Constraints
– castle will contain between 1 and 50 elements, inclusive.
– Each element of castle will contain between 1 and 50 characters, inclusive.
– All elements of castle will contain the same number of characters.
– Each character of each element of castle will be either ‘.’ or uppercase ‘X’.

Examples
0)

{ “….”,
“….”,
“….”,
“….” }

Returns: 4

Here we can place 4 guards on one of the diagonals and that will satisfy us.
1)

{ “XX…”,
“.XX..”,
“…XX” }

Returns: 0

No additional guards needed.
2)

{ “….XXXX”,
“……..”,
“XX.X.XX.”,
“……..”,
“……..” }

Returns: 3

3)

{ “……..X..”,
“…X…….”,
“………..”,
“..X…X….”,
“………..”,
“………..”,
“……..X..”,
“………..”,
“………..”,
“……..X..”,
“…..X…..” }

Returns: 6

This code has not yet been tested,
and hence it cannot be guaranteed that this code is a proper solution to the above problem.
Watch the space after some span of time and this will be done.



#include "iostream.h"
#include "conio.h"

char Map[50][50];
int maxE=0, maxC=0;

int scan(int X, int Y, char* dir){
 int i=0;
 if(dir == "horiz"){
  for(i=0;i< maxC;i++){
   if(Map[i][Y]=='X')
    return 0;
  }
  return 1;
 }
 if(dir == "vert"){
  for(i=0;i< maxE;i++){
   if(Map[X][i]=='X')
    return 0;
  }
  return 1;
 }
}

int findShortage(int seedC, int seedE){
 static int less=0;
 int horiz, vert;
 if(seedE > maxE){
  return less;
 }
 horiz = scan(seedC,seedE,"horiz");
 vert  = scan(seedC,seedE,"vert");
 if(horiz && !vert){
  findShortage(seedC+1, seedE);
 }
 if(vert && !horiz){
  findShortage(seedC,seedE+1);
 }
 if(horiz && vert ){
  less++;
  findShortage(seedC+1, seedC+1);
 }
}

void main(void){
 clrscr();
 int i,j;
 cout<>maxE;
 cout<>maxC;
 for(i=0;i< maxE;i++)
  for(j=0;j< maxC;j++){
   cout<<"Map["<< i <<"]["<< j <<"]";
   cin>>Map[i][j];
  }

 cout<< findShortage(0,0);
 getch();
}

This Programming Challenge requires you to read in a text file, process the contents of the file and output the results.
The file consists of 50 lines of text. Each line consists of 3 sections: First is 3 characters made up of spaces and digits that you can ignore, then 80 characters which are either a space or a + then another 2 characters. The first 3 and last 2 characters are just 1 or 2 digits line numbers padded with spaces. The important part is the 80 characters.
The file would look something like this…

File starts here:
1 1
27 ++++++++ ++++ +++++++ ++++++++++++ +++ 27
28 ++++++++ ++++ +++++++ ++++++++++++ + ++++ ++++++ 28

This is a map of a game world where each location is either land or sea.Land points are +, sea are spaces. What you have to do is process the 80 x 50 map, determine how many continents there are and output how many continents there are plus a list of those continents, together with the size of each continent. The definition of a continent is one or more connected locations where each location is a +. Connected means that in a 3 x 3 grid around a +, if any of the 8 surrounding locations contains a + then the two are connected.

the map file can be downloaded from this location.

I present the code for this problem as below.

/* C++ */

/*
author  : pranav prakash
country : india
email   : pranavprakash@programmer.net
*/

#include 

char map[50][85];

int getMap(char*);
int scanContinent(int, int , int);
int findNext();

int getMap(char* mapfile){
 fstream fl;
 char v;
 fl.open(mapfile, ios::in);
 if(fl.bad() || fl.eof()){
  return -1;
 }
 for(int i=0; i<50; i++){
  for(int j=0; j<85; j++){
   fl.read((char*)&v, sizeof(v));
   map[i][j] = v;
  }
 }
 return 1;
}

int findNext(){
 int Xpos=0, Ypos=0, unit=0, i, j;
 static int continent=1;

 for(i=0; i<50; i++)
  for(j=3; j<83; j++)
   if(map[i][j] == '+'){
    Xpos = i;
    Ypos = j;
    cout<<"\n\n--- Continent Number: "<< continent<<" ---";
    //cout<<"\nXpos: "<< Xpos<<" Ypos: "<< Ypos<<"\n";
    unit = scanContinent(Xpos, Ypos, continent++);
    cout<<"\nNo. of units: "<< unit;

   }

 return 0; // No more continents left in the map;

}

int scanContinent(int seedX, int seedY, int continent){
 static int count=0;
 if(map[seedX][seedY] == '+'){
  count++;
  map[seedX][seedY] = continent;

  scanContinent(seedX,seedY+1,continent);
  scanContinent(seedX,seedY-1,continent);

  scanContinent(seedX+1,seedY+1,continent);
  scanContinent(seedX+1,seedY,continent);
  scanContinent(seedX+1,seedY-1,continent);

  scanContinent(seedX-1,seedY+1,continent);
  scanContinent(seedX-1,seedY,continent);
  scanContinent(seedX-1,seedY-1,continent);
 }
 return count;
}

void main(void){
 if(getMap("map.txt")){
  findNext();
 }
}

The original problem statement can be viewed at this location.

This solution was actually selected as a correct solution and was in the top 20 codes. The actual content can be seen at this location.

1. Write a “Hello World” program in ‘C’ without using a semicolon.

void main()
{
  if(printf("Hello World\n"))
  {
  }
}

2. Write a C++ program without using any loop (if, for, while etc) to print numbers from 1 to 100 and 100 to 1;

I haven’t tried this code, but i think that it will work, please check and tell me…

#include 

int one_round = 0;

int loop(int a){
    (one_round == 0)?printf("%d\n", a++):printf("%d\n", a--);
    (a == 100)? one_round++: one_round;
    (a==0)?1:loop(a);
}

int main(void){
    loop(1);
    return 0;
}

3. C/C++ : Exchange two numbers without using a temporary variable.
The code could be like this …

swap(int a, int b) {
    a = a+b;
    b = a-b;
    a = a-b;
}

4. C/C++ : Find if the given number is a power of 2.

Using the code given in Problem statement 10, or otherwise, convert the number
into binary, if that binary number has only one 1 and that too, on the extreme left,
then the corresponding decimal number is a power of two. However, this process doesn't appeal
much to me. So i am thinking of a better and optimal way.

5. C/C++ : Multiply x by 7 without using multiplication (*) operator.

int mult(int multiplicand, int multiplier) {
   int result=1;
   for( multiplicand; multiplicand >0; multiplicand -- ) {
       result = result + multiplier;
   }
   return result;
}

6. C/C++ : Write a function in different ways that will return f(7) = 4 and f(4) = 7

This is a simple mathematical problem regarding basic calculus and geometry.

Lets see
f(7) = 4
f(4) = 7
hence, [f(7)-f(4)] / [7-4] = -1
It simply means that the function is a straight line with slope, m = -1.
So let the equation be y=mx+c in slope intercept form.
Now after doing little calculations, we can find that c = 11.
Hence the required function is
 f(x) + x = 11

7. Remove duplicates in array

The code for this goes as follows.

int removeDuplicates(int arr[], int len) {
   int Dup;
   for( int i=0; i < len; i++){
      Dup = arr[i];
      for( int j = i; j < len; j++) {
         if( Dup == arr[j]) { // Duplicate entry detected
            for( int k=j; k < len; k++) {
                 arr[k] = ar[k+1];
            }
      }
   }
   return 0;
}

8. Finding if there is any loop inside linked list.

class Node{
  int data;
  int traversed;
  struct Node *next;

  Node(){
    data = 9999;
    traversed = 0;
  }
};

typedef Node* Nodeptr;

Now traverse the Linked List, and keep on marking the data item traversed as 1.
If in the course of traversing, you arrive at a node that has already been traversed,
i.e. its traversed == 1, then there is a loop present in the List.

9. Remove duplicates in an no key access database without using an array

10. Convert (integer) number in binary without loops.

int int2bin(int n) {
   static int bin[8], i;
   i = -1;
   i++;
   int a1 = n % 2 ;
   bin[i] = a1;
   int a2 = n / 2;
   if( a2 == 0)
       return 0;
   int2bin(a2);
}

11. Write a program whose printed output is an exact copy of the source. Needless to say, merely
echoing the actual source file is not allowed.

Such programmes are called Quines. The various possible ways for coding them are given ::

char *f="char *f=%c%s%c;%c#define Q '%c'%c#define N '%cn'%c#define B '%c%c'%c#include 
%cvoid main(){printf(f,Q,f,Q,N,Q,N,B,N,B,B,N,N,N);}%c";
#define Q '"'
#define N '\n'
#define B '\\'
#include 
void main(){printf(f,Q,f,Q,N,Q,N,B,N,B,B,N,N,N);}

Another method that can be used is as follows::


#include 
void main(void) {
   fstream fl;
   char a;
   fl.open("quine.cpp");
   while( !fl.eof()){
   fl.read((char*)&a, sizeof(a);
     cout<< a;
   }
   fl.close();
}

Save this program as "quine.cpp"

12. From a ‘pool’ of numbers (four ‘1’s, four ‘2’s …. four ‘6’s), each player selects a number and adds it to the total. Once a number is used, it must be removed from the pool. The winner is the person whose number makes the total equal 31 exactly.

13. Swap two numbers without using a third variable.

swap(int a, int b) {
    a = a+b;
    b = a-b;
    a = a-b;
}

14. Given an array (group) of numbers write all the possible sub groups of this group.

If you have any ambiguity or clarifications, please mail me….