prestoring and fetching

why hashing?

1. if number
2. for i<n
3. arr[i} == number
4. cnt = cnt+1; -^
5. return cnt

now, hashing is → prestoring/ fetching

1. a array of size 13

2. with initial as 0; it is a hash array

   

   

3. updating with the iteration

code

1. int hash[13];
2. for(int i =0; i<n;i++){
3. hash[arr[i]]++;
4. }
7. cout << hash[number] <<endl;

one imp thing for 10^7 will not go for the array

1. if array it will be max 10^6

1. can we do character hashing?

   1. yes
   2. Q→ quries
      1. a→1
      2. b→2
      3. c→3

    ![Untitled](<https://prod-files-secure.s3.us-west-2.amazonaws.com/94b73934-74f9-494c-9e82-21f4a31e6ebf/66777cb8-658d-485d-ac2a-fdc72e51f587/Untitled.png>)
   

2. aperently the formula will be character - ‘a’

 then, e-a =4,
 
 vagera vagera

problem 2- code

1. int hash[26] = {0};
3. for(int) s.size(); i++
   1. hash[s[i]-’a’] ++;

always prefer arrays

number hashing → stc/collection

stl comes here: vahi ki 10^9 jitna isme nahi aa paye ga

1. map and unordered map,

   1. map

      1. arr= 1,2,3,1,3,2

         

      2. code:

         1. map<long or int,int> mpp;

         2. for

            1. map[arr[i]]++

         3. itrate in map:

            1. for(auto it : mpp){
               1. it.first, it.second

2. HashMap

   1. store all thing in sorted order:
   2. auto I : map
   3. map<char,int>
      1. mpp[s[i]]++;
   4. map: log(n)

3. Unordered_map

   1. no perticulat order

      1. advantage: o(1);

   2. worst case: o(n);

   3. first case : take unordered_map → if tle, then go to normal map