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]. 


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;

	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;
		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


	return is_doable();

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

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

	char buffer[10];

	for (int j = 0; j < m; ++j)
		scanf("%s", 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;
    for(size_t i = 0; i < M; ++i)
        cin >> str;
    for(const auto& word : ransom)
        auto found = magazine.find(word);
        if(found == magazine.end())
            cout << "No";
            return 0;
    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;
        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
        Solution s = new Solution(scanner.nextLine(), scanner.nextLine());

                        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(' ')
answer = ransom_note(magazine, ransom)

View More Similar Problems

Square-Ten Tree

The square-ten tree decomposition of an array is defined as follows: The lowest () level of the square-ten tree consists of single array elements in their natural order. The level (starting from ) of the square-ten tree consists of subsequent array subsegments of length in their natural order. Thus, the level contains subsegments of length , the level contains subsegments of length , the

View Solution →

Balanced Forest

Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a

View Solution →

Jenny's Subtrees

Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .

View Solution →

Tree Coordinates

We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For

View Solution →

Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

View Solution →

Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ

View Solution →