c++ stl map

Hello everyone!

This time I am going to tell you about  MAP -> a very very important stuff that saves lot of your time during a coding contest .
Just like stack , queue ,      map is also a container class of STL .

A map contains two values , one of them is the key and other is the value stored at that key . A particular value can be accessed or modified by using its key . A key of a value is unique i.e map has unique keys which can not be modified .
following example will clarify everything :

#include<iostream>
#include<string.h>
#include<string>
#include<map>

using namespace std;

int main()
{
     map<string,int>employees;   // here the first one(string) is the key and int is the value stored at
                                                   //at that key
     employees.insert(pair<string,int>("vikram",1932)) ; //inserting values into map
      employees.insert(pair<string,int>("Rahul" ,1231)) ;
     employees.insert(pair<string,int>("arjun" ,1012)) ;
 
     cout<<"map size : "<<employees.size()<<endl; // to get  the size of map
     // iterating thorugh a map
    // In STL iterators provide a means for accessing data stored in container classes such a vector
   //  list , map etc .
    // declaring an iterator ->  class_name<template_parameters>::iterator name
   map<string,int>::iterator it = employees.begin(); // referencing iterator to the beginning of map
   for(it; it!= employees.end() ; it++ )
   {
         cout<< it->first<<"  "<<it->second<<endl;  // first gives  you the key ,, second gives you value
   }cout<<endl;
    //iterating the map in reverse manner
 
    map<string,int>::reverse_iterator  bt =  employees.rbegin();
   for(bt ;  bt!= employees.rend(); bt++)
   {
         cout<< bt->first<<"  "<<bt->second<<endl;
   }



   /*  to check whether a key is present in the map */
   // suppose we want to check  if  a key named "arjun" is present in the map or not
   // for this we use map.count(key_name) ;  it returns value >0 if that key is already in map else 0

   if( employees.count("arjun")>0){cout<<"this key is present "<<endl; }
 
      /*  accessing/modifying  value stored at a particular key*/
      // accessing :
      cout<<"value stored at arjun : "<<employees.at("arjun")<<endl;
     //modifying :
      employees.at("arjun") = 500;
    // inserting new key using [ ] :
     employees ["tata"] = 123;
      cout<<"value stored at tata : "<<employees.at("tata")<<endl;
 
 }

output is :
map size : 3
Rahul  1231
arjun  1012
vikram  1932

vikram  1932
arjun  1012
Rahul  1231
this key is present
this key is present
value stored at arjun : 1012
value stored at tata : 123

  

you can see that value stored in a map is already sorted . Time complexity of std:: map is of log(n)

I will be writing about tie complexity in another post .

If regarding map you have any doubt , visit this page this one

Here are some questions which can be solved using map . Please go through this link  if you hate Ramsay Bolton and enjoyed the Battle of Bastards :)