Using Basic Vectors

VECTORS [STL :C++]

Want to sort an array of numbers but failing to choose a perfect sorting algorithm for the same looks cumbersome ? Want to optimize your solutions on sites like Codechef , HackerRank ? 

Then my friend C++ has the answer for you stored in it's STL library. 

STL library contains Vectors Container which is just like an array but with better functionalities. I have commented in the code the syntax and declaration of vector for easy understanding.

Question :

Given the list of numbers, you are to sort them in non decreasing order.

Input

t – the number of numbers in list, then t lines follow [t <= 10^6].

Each line contains one integer: N [0 <= N <= 10^6]

Output

Output given numbers in non decreasing order.

Example

Input:

5
5
3
6
7
1

Output:

1
3
5
6
7

Answer : 

 #include <iostream>  
 #include <vector>  //for vector  
 #include <algorithm> //for sort  
 #include <stdio.h>  
 using namespace std;  
 int main()  
 {  
   long int t,i,temp;  
   vector<long int> v;      //Creating a Vector v  
   scanf("%ld",&t);  
   for(i=0;i<t;i++)  
   {  
     scanf("%ld",&temp);  
     v.push_back(temp);    //adding the element at the end of the vector  
   }  
   sort(v.begin(),v.end());    //the function "sort" is defined in <algorithm.h>  
                    //It takes parameters from the vector as to range which it has to sort  
   for(i=0;i<t;i++)  
   {  
     printf("%ld\n",v[i]);  
   }  
   return 0;  
 }  

IEEE XTREME 7.0

Solution Facebook Hacker Cup 2013

Square Detector

You want to write an image detection system that is able to recognize different geometric shapes. In the first version of the system you settled with just being able to detect filled squares on a grid.

You are given a grid of N×N square cells. Each cell is either white or black. Your task is to detect whether all the black cells form a square shape.

Input

The first line of the input consists of a single number T, the number of test cases.

Each test case starts with a line containing a single integer N. Each of the subsequent N lines contain N characters. Each character is either "." symbolizing a white cell, or "#" symbolizing a black cell. Every test case contains at least one black cell.

Output

For each test case i numbered from 1 to T, output "Case #i: ", followed by YES or NO depending on whether or not all the black cells form a completely filled square with edges parallel to the grid of cells.

Constraints

1 ≤ T ≤ 20

1 ≤ N ≤ 20

Example

Test cases 1 and 5 represent valid squares. Case 2 has an extra cell that is outside of the square. Case 3 shows a square not filled inside. And case 4 is a rectangle but not a square.

Example input

5
4
..##
..##
....
....
4
..##
..##
#...
....
4
####
#..#
#..#
####
5
#####
#####
#####
#####
.....
5
#####
#####
#####
#####
#####

Example output 

Case #1: YES
Case #2: NO
Case #3: NO
Case #4: NO
Case #5: YES



Note: Just omit the last line in your outpfile cause that is extra and your file won't be 
accepted.


SOLUTION :
 #include <iostream>  
 #include <vector>  
 #include <algorithm>  
 #include <fstream>  
 using namespace std;  
 int main()  
 {  
   //ifstream in("in.txt");  
   //ofstream out("out.txt");  
   char a[21][21];  
   int t,num=0,n;  
   cin>>t;  
   //INPUT FROM THE USER  
   while(t--)  
   {  
     num++;  
     int i,j,flag=0,k,x,count=0,final_count=0;  
           vector<int> y;  
     cin>>n;  
     for(i=1;i<=n;i++)  
   {  
     for(j=1;j<=n;j++)  
     {  
       cin>>a[i][j];  
       if (a[i][j]=='#')  
       {  
         final_count++;  
       }  
     }  
   }  
   //STORING FIRST ROW WITH THE OCCURENCE OF "#"  
   for(i=1;i<=n;i++)  
   {  
     for(j=1;j<=n;j++)  
     {  
       if (a[i][j]=='#')  
       {  
           flag=1;  
           x=i;  
           y.push_back(j);  
           //cout<<i<<","<<j<<endl;  
       }  
     }  
     if(flag==1)  
     {  
       break;  
     }  
   }  
   int z=0;  
   for(i=1;i<=n;i++)  
   {  
     for(j=1;j<=n;j++)  
     {  
       if (a[i][j]=='#')  
       {  
         for(k=0;k<y.size();k++)  
         {  
           if(j==y[k])  
           {  
             count++;  
           }  
         }  
       }  
     }  
     if(count==y.size())  
     {  
         //cout<<"row "<<i<<"matched"<<endl;  
         if(x==i)  
         {  
           z++;  
           x++;  
         }  
     }  
     count=0;  
   }  
      //cout<<"z "<<z <<" y.size() "<<y.size()<<endl;  
   cout<<"Case #"<<num;  
   if(final_count!=(z*z))  
   {  
     cout<<": NO"<<endl;  
   }  
   else if(z==y.size())  
   {  
     cout<<": YES"<<endl;  
   }  
   else  
   {  
     cout<<": NO"<<endl;  
   }  
   }  
 return 0;  
 }  




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