Maps-STL C++


Problem Statement :


Maps are a part of the C++ STL.Maps are associative containers that store elements formed by a combination of a key value and a mapped value, following a specific order.The mainly used member functions of maps are:

    Map Template:

    std::map <key_type, data_type>

    Declaration:

    map<string,int>m; //Creates a map m where key_type is of type string and data_type is of type int.

    Size:

    int length=m.size(); //Gives the size of the map.

    Insert:

    m.insert(make_pair("hello",9)); //Here the pair is inserted into the map where the key is "hello" and the value associated with it is 9.

    Erasing an element:

    m.erase(val); //Erases the pair from the map where the key_type is val.

    Finding an element:

    map<string,int>::iterator itr=m.find(val); //Gives the iterator to the element val if it is found otherwise returns m.end() .
    Ex: map<string,int>::iterator itr=m.find("Maps"); //If Maps is not present as the key value then itr==m.end().

    Accessing the value stored in the key:

    To get the value stored of the key "MAPS" we can do m["MAPS"] or we can get the iterator using the find function and then by itr->second we can access the value.

To know more about maps click Here.

You are appointed as the assistant to a teacher in a school and she is correcting the answer sheets of the students.Each student can have multiple answer sheets.So the teacher has Q queries:

1 X Y :Add the marks  Y to the student whose name is X
2 X: Erase the marks of the students whose name is X
3 X: Print the marks of the students whose name is X . (If X didn't get any marks print  0.)

Input Format

The first line of the input contains Q where Q is the number of queries. The next Q lines  contain 1 query each.The first integer type, of each query is the type of the query.If query is of type 1 , it consists of one string and an integer X and Y where is  X the name of the student and Y is the marks of the student.If query is of type 2 or 3 ,it consists of a single string  X where X is the name of the student. 

Constraints

  1 <=  Q  <= 10^5
  1 <= type  <= 3
  1 <=  | X |   <=  6
  1 <=  Y    <= 10^3

Output Format

For queries of type 3 print the marks of the given student.


Solution :



title-img 


                            Solution in C :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <set>
#include <map>
#include <algorithm>
using namespace std;


int main() {
    int n;
    cin >> n;
    
    map<string, int> m;
    
    while (n--) {
        int t;
        cin >> t;
        
        string s;
        cin >> s;
        
        if (t == 1) {
            int a;
            cin >> a;
            
            m[s] += a;
        } else if (t == 2) {
            m[s] = 0;
        } else {
            cout << m[s] << "\n";
        }
    }
    
    return 0;
}








View More Similar Problems

Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

View Solution →

Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

View Solution →

Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

View Solution →

Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

View Solution →

Binary Search Tree : Lowest Common Ancestor

You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b

View Solution →

Swap Nodes [Algo]

A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from

View Solution →