# Hash Tables: Ransom Note

### Problem Statement :

```Harold is a kidnapper who wrote a ransom note, but now he is worried it will be traced back to him through his handwriting. He found a magazine and wants to know if he can cut out whole words from it and use them to create an untraceable replica of his ransom note. The words in his note are case-sensitive and he must use only whole words available in the magazine. He cannot use substrings or concatenation to create the words he needs.

Given the words in the magazine and the words in the ransom note, print Yes if he can replicate his ransom note exactly using whole words from the magazine; otherwise, print No.

For example, the note is "Attack at dawn". The magazine contains only "attack at dawn". The magazine has all the right words, but there's a case mismatch. The answer is No.

Function Description

Complete the checkMagazine function in the editor below. It must print Yes if the note can be formed using the magazine, or No

Input Format

The first line contains two space-separated integers m, and n , the numbers of words in the and the magazine and the note.
The second line contains m space-separated strings, each magazine[i].
The third line contains n space-separated strings, each note[i].

Constraints

1 <= m,  n <= 300
1 <= | magazine[i] | , | note[i] | <= 5

.
Each word consists of English alphabetic letters (i.e, a to z and A to Z).

.Output Format

Print Yes if he can use the magazine to create an untraceable replica of his ransom note. Otherwise, print No.

checkMagazine has the following parameters:

magazine: an array of strings, each a word in the magazine
note: an array of strings, each a word in the ransom note

.```

### Solution :

```                            ```Solution in C :

In C :

#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#include <string.h>

#define DICT_SIZE 50007

typedef struct _dict_entry
{
uint64_t hash;
char *str;
int value;
} *dict_entry;

dict_entry dict[DICT_SIZE];

uint64_t get_hash(char *s)
{
int shift = 0;

uint64_t hash = 0;
while (*s != '\0')
{
hash |= *s << shift;
shift += 7;
++s;
}

return hash;
}

dict_entry dict_get(char *s)
{
uint64_t hash = get_hash(s);
size_t pos = hash % DICT_SIZE;
while (dict[pos] != NULL && dict[pos]->hash != hash)
pos = (pos + 1) % DICT_SIZE;
return dict[pos];
}

void dict_inc(char *s)
{
dict_entry entry = dict_get(s);

if (entry)
{
entry->value += 1;
}
else
{
entry = (dict_entry)malloc(sizeof(struct _dict_entry));
memset(entry, 0, sizeof(struct _dict_entry));

entry->hash = get_hash(s);
entry->str = strdup(s);
entry->value = 1;

size_t pos = entry->hash % DICT_SIZE;
while (dict[pos] != NULL)
pos = (pos + 1) % DICT_SIZE;
dict[pos] = entry;
}
}

int is_doable()
{
for (int j = 0; j < DICT_SIZE; ++j)
if (dict[j] && dict[j]->value < 0)
return 0;
return 1;
}

int solve(int n)
{
char buffer[10];

for (int j = 0; j < n; ++j)
{
scanf("%s", buffer);

dict_entry entry = dict_get(buffer);
if (entry == NULL)
return 0; // unbek. wort

entry->value--;
}

return is_doable();
}

int main()
{
#ifdef _DEBUG
char FNAME[250];
strcpy(FNAME, __FILE__);
strcpy(strchr(FNAME, '.'), ".txt");
freopen(FNAME, "rt", stdin);
#endif

int m, n;
scanf("%d %d", &m, &n);

char buffer[10];

for (int j = 0; j < m; ++j)
{
scanf("%s", buffer);
dict_inc(buffer);
}

printf(solve(n) ? "Yes" : "No");
}```
```

```                        ```Solution in C++ :

In C++ :

#include <string>
#include <iostream>
#include <unordered_set>

using namespace std;

unordered_multiset<string> magazine, ransom;

int main()
{
size_t N, M;
cin >> N >> M;

if(N < M)
{
cout << "No";
return 0;
}

string str;

for(size_t i = 0; i < N; ++i)
{
cin >> str;
magazine.insert(str);
}

for(size_t i = 0; i < M; ++i)
{
cin >> str;
ransom.insert(str);
}

for(const auto& word : ransom)
{
auto found = magazine.find(word);

if(found == magazine.end())
{
cout << "No";
return 0;
}

magazine.erase(found);
}
cout << "Yes";
}```
```

```                        ```Solution in Java :

In java :

import java.util.*;

public class Solution {
Map<String, Integer> magazineMap;
Map<String, Integer> noteMap;

public Solution(String magazine, String note) {
this.noteMap = new HashMap<String, Integer>();
this.magazineMap = new HashMap<String, Integer>();

Integer occurrences;

for(String s : magazine.split("[^a-zA-Z]+")) {
occurrences = magazineMap.get(s);

if(occurrences == null) {
magazineMap.put(s, 1);
}
else {
magazineMap.put(s, occurrences + 1);
}
}

for(String s : note.split("[^a-zA-Z]+")) {
occurrences = noteMap.get(s);

if(occurrences == null) {
noteMap.put(s, 1);
}
else {
noteMap.put(s, occurrences + 1);
}
}

}

public void solve() {
boolean canReplicate = true;
for(String s : noteMap.keySet()) {
if(!magazineMap.containsKey(s) || (magazineMap.get(s) < noteMap.get(s)) ) {
canReplicate = false;
break;
}
}

System.out.println( (canReplicate) ? "Yes" : "No" );
}

public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int m = scanner.nextInt();

// Eat whitespace to beginning of next line
scanner.nextLine();

Solution s = new Solution(scanner.nextLine(), scanner.nextLine());
scanner.close();

s.solve();
}
}```
```

```                        ```Solution in Python :

In Python3 :

def ransom_note(magazine, ransom):
rc = {} # dict of word: count of that word in the note
for word in ransom:
if word not in rc:
rc[word] = 0
rc[word] += 1

for word in magazine:
if word in rc:
rc[word] -= 1
if rc[word] == 0:
del rc[word]
if not rc:
return True
return False

m, n = map(int, input().strip().split(' '))
magazine = input().strip().split(' ')
ransom = input().strip().split(' ')
print("Yes")
else:
print("No")```
```

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your \$inOrder* func

## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

## 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.

## 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

## 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 <

## 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