Notes

First What we can understand from the question is

We are given a set of characters which can be used any number of times
and we are given a word/sentence which we have to tell if we can form using those given set of words or not..

n- length of our magzine word
m - length of ransome note
1st Approach :- Bruteforce ( n*m )
We need to find if we can form ransome note from magzine , lets iterate whole magzine for each alphabet in ransome note

Code :-
for( int i =0;i<ransomenote.size();i++)
{
bool found = false;
for( int j = 0; j<magzine.size() ; j++)
if(ransomenote[i] == magzine[j] )
flag=true;
if(!flag)
return false;
}
return true;

We just saw that we here complexity is n*m which will give us TLE ...

2. Binary Search ( nlogn)
what we did in bruteforce was finding each char linerarly now we can do binary search by sorting the magzine
string....
Try it urself ..... ( Use STL binary search fn )

3. Lets just create a map of all the characters of the magzine string

Code :-
map<char,bool>mp ;
for( auto i:magzine)
mp[i]=true;
for(auto i:ransomenote)
if(!mp[i])
return false;
return true;

4. Using Vector as a map

Note : in question this is mentioned "ransomNote and magazine consist of lowercase English letters"
which means we actually just need an array of size 26 . and one more thing is that 'a'-'a' = 0 , 'b'-'a'=1 _____'z'-'a'=25 , this will act as our index

Lets Code it now
vector<int>Map(26,0) ; // 0 is to initialise vector with 0 initially
for(auto i:magzine)
Map[i-'a']++;
for(auto i:ransomenote)
if(!Map[i-'a'])
return false;
return true;