MostFrequentSubstring , Frekventen String AIPS

Фреквентен стринг Problem 3 (0 / 1)

Даден е стринг. Треба да се најде најчестиот под-стринг кој е дел од него и да се испечати. Доколку два под-стринга се исто фреквентни, тогаш се печати подолгиот. Доколку и овој услов го исполнуваат тогаш се печати лексикографски помалиот.

Пример: За стрингот "abc" под-стрингови се "a", "b", "c", "ab", "bc", "abc". Сите имаат иста честота па затоа се печати најдолгиот "abc".

Име на класата (Java): MostFrequentSubstring.


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

class MapEntry<K extends Comparable<K>,E> implements Comparable<K> {

    K key;
    E value;

    public MapEntry (K key, E val) {
        this.key = key;
        this.value = val;
    }

    public int compareTo (K that) {
        @SuppressWarnings("unchecked")
        MapEntry<K,E> other = (MapEntry<K,E>) that;
        return this.key.compareTo(other.key);
    }

    public String toString () {
        return "(" + key + "," + value + ")";
    }
}

class SLLNode<E> {
    protected E element;
    protected SLLNode<E> succ;

    public SLLNode(E elem, SLLNode<E> succ) {
        this.element = elem;
        this.succ = succ;
    }

    @Override
    public String toString() {
        return element.toString();
    }
}

class CBHT<K extends Comparable<K>, E> {

    private SLLNode<MapEntry<K,E>>[] buckets;

    @SuppressWarnings("unchecked")
    public CBHT(int m) {
        buckets = (SLLNode<MapEntry<K,E>>[]) new SLLNode[m];
    }

    private int hash(K key) {
        return Math.abs(key.hashCode()) % buckets.length;
    }

    public SLLNode<MapEntry<K,E>> search(K targetKey) {
        int b = hash(targetKey);
        for (SLLNode<MapEntry<K,E>> curr = buckets[b]; curr != null; curr = curr.succ) {
            if (targetKey.equals(((MapEntry<K, E>) curr.element).key))
                return curr;
        }
        return null;
    }

    public void insert(K key, E val) {      // Insert the entry <key, val> into this CBHT.
        MapEntry<K, E> newEntry = new MapEntry<K, E>(key, val);
        int b = hash(key);
        for (SLLNode<MapEntry<K,E>> curr = buckets[b]; curr != null; curr = curr.succ) {
            if (key.equals(((MapEntry<K, E>) curr.element).key)) {
                curr.element = newEntry;
                return;
            }
        }
        buckets[b] = new SLLNode<MapEntry<K,E>>(newEntry, buckets[b]);
    }

    public void delete(K key) {
        int b = hash(key);
        for (SLLNode<MapEntry<K,E>> pred = null, curr = buckets[b]; curr != null; pred = curr, curr = curr.succ) {
            if (key.equals(((MapEntry<K,E>) curr.element).key)) {
                if (pred == null)
                    buckets[b] = curr.succ;
                else
                    pred.succ = curr.succ;
                return;
            }
        }
    }

    public String toString() {
        String temp = "";
        for (int i = 0; i < buckets.length; i++) {
            temp += i + ":";
            for (SLLNode<MapEntry<K,E>> curr = buckets[i]; curr != null; curr = curr.succ) {
                temp += curr.element.toString() + " ";
            }
            temp += "\n";
        }
        return temp;
    }

}

public class MostFrequentSubstring {
    public static void main (String[] args) throws IOException {
        CBHT<String,Integer> tabela = new CBHT<String,Integer>(300);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        String word = br.readLine().trim();

        // Vashiot kod tuka....
        char []letters = word.toCharArray();

        // generates substring

        for( int i=1; i < letters.length; i++ ) { // determines substring lenght

            for (int j = 0; j < letters.length - i + 1; j++) { // determines substring beginning and puts them in hash

                String str = "";
                for (int k = j; k < j + i; k++) { // determines substrings

                    str += Character.toString(letters[k]);
                }

                SLLNode<MapEntry<String, Integer>> node = tabela.search(str); // determines if the substring has been allready hashed

                if (node == null)
                    tabela.insert(str, 1);

                else
                    tabela.insert(str, (node.element.value + 1)); // enlarges the sub substring frequency by 1
            }
        } // done with hashing

        // looking for the most frequent string in the hash
        String mostFrequent = "";
        int frequencyMax = 0;

        for( int i=1; i < letters.length; i++ ) { // generates substrings, same as above

            for (int j = 0; j < letters.length - i + 1; j++) {

                String str = "";
                for (int k = j; k < j + i; k++) {

                    str += Character.toString(letters[k]);
                }

                SLLNode<MapEntry<String, Integer>> node = tabela.search(str);

                if (node == null)
                    continue;

                else {
                    if ( node.element.value > frequencyMax ) {
                        frequencyMax = node.element.value;
                        mostFrequent = str;
                    }
                    else if( node.element.value == frequencyMax ) {
                        if( str.length() > mostFrequent.length() ) {
                            frequencyMax = node.element.value;
                            mostFrequent = str;
                        }
                        else if( str.length() == mostFrequent.length() ) {
                            if( lexikographicValue( str ) < lexikographicValue( mostFrequent ) ) {
                                frequencyMax = node.element.value;
                                mostFrequent = str;
                            }
                        }
                    }
                }
            }
        }

        System.out.println( mostFrequent );
    }

    public static int lexikographicValue( String str ) {

        char letters[] = str.toCharArray();
        int val = 0;
        for( int i = 0; i < letters.length; i++ ) {
            val += (int)letters[i];
        }
        return val;
    }
}