Sunday, July 31, 2016

Multi Search - Cracking Coding Interview


Given a string band an array of smaller strings T, design a method to search b for each small string in T.

给一个字符串S和一个字符串数组T(T中的字符串要比S短许多),设计一个算法, 在字符串S中查找T中的字符串。
后缀Trie的查找效率很优秀,如果你要查找一个长度为n的字符串,只需要O(n)的时间, 比较次数就是字符串的长度,相当给力。 但是,构造字符串S的后缀Trie却需要O(m^2 )的时间, (m为S的长度),及O(m^2 )的空间。

Alternatively, we can add all the smaller strings into a trie
public static Trie createTreeFromStrings(String[] smalls, int maxSize) {
Trie tree = new Trie();
for (String s : smalls) {
if (s.length() <= maxSize) {
tree.insertString(s, 0);
}
}
return tree;
}

public static ArrayList<String> findStringsAtLoc(TrieNode root, String big, int start) {
ArrayList<String> strings = new ArrayList<String>();
int index = start;
while (index < big.length()) {
root = root.getChild(big.charAt(index));
if (root == null) break;
if (root.terminates()) {
strings.add(big.substring(start, index + 1));
}
index++;
}
return strings;
}

public static void insertIntoHashMap(ArrayList<String> strings, HashMapList<String, Integer> map, int index) {
for (String s : strings) {
map.put(s, index);
}
}

public static HashMapList<String, Integer> searchAll(String big, String[] smalls) {
HashMapList<String, Integer> lookup = new HashMapList<String, Integer>();
TrieNode root = createTreeFromStrings(smalls, big.length()).getRoot();
for (int i = 0; i < big.length(); i++) {
ArrayList<String> strings = findStringsAtLoc(root, big, i);
insertIntoHashMap(strings, lookup, i);
}
return lookup;
}

public static void subtractValue(ArrayList<Integer> locations, int delta) {
if (locations == null) return;
for (int i = 0; i < locations.size(); i++) {
locations.set(i, locations.get(i) - delta);
}
}

public static Trie createTrieFromString(String s) {
Trie trie = new Trie();
for (int i = 0; i < s.length(); i++) {
String suffix = s.substring(i);
trie.insertString(suffix, i);
}
return trie;
}

public static HashMapList<String, Integer> searchAll(String big, String[] smalls) {
HashMapList<String, Integer> lookup = new HashMapList<String, Integer>();
Trie tree = createTrieFromString(big);
for (String s : smalls) {
/* Get terminating location of each occurrence.*/
ArrayList<Integer> locations = tree.search(s);
/* Adjust to starting location. */
subtractValue(locations, s.length());
/* Insert. */
lookup.put(s, locations);
}
return lookup;
}
public class Trie {
private TrieNode root = new TrieNode();
public ArrayList<Integer> search(String s) {
return root.search(s);
}
public void insertString(String str, int location) {
root.insertString(str, location);
}
public TrieNode getRoot() {
return root;
}

public class TrieNode {
private HashMap<Character, TrieNode> children;
private ArrayList<Integer> indexes;
public TrieNode() { 
children = new HashMap<Character, TrieNode>();
indexes = new ArrayList<Integer>();
}
public void insertString(String s, int index) {
if (s == null) return;
indexes.add(index);
if (s.length() > 0) {
char value = s.charAt(0);
TrieNode child = null;
if (children.containsKey(value)) {
child = children.get(value);
} else {
child = new TrieNode();
children.put(value, child);
}
String remainder = s.substring(1);
child.insertString(remainder, index + 1);
} else {
children.put('\0', null);
}
}
public ArrayList<Integer> search(String s) {
if (s == null || s.length() == 0) {
return indexes;
} else {
char first = s.charAt(0);
if (children.containsKey(first)) {
String remainder = s.substring(1);
return children.get(first).search(remainder);
}
}
return null;
}
public boolean terminates() {
return children.containsKey('\0');
}
public TrieNode getChild(char c) {
return children.get(c);
}
}


public static boolean isSubstringAtLocation(String big, String small, int offset) {
  for (int i = 0; i < small.length(); i++) {
    if (big.charAt(offset + i) != small.charAt(i)) {
      return false;
    }
  }
  return true;
}

public static ArrayList<Integer> search(String big, String small) {
  ArrayList<Integer> locations = new ArrayList<Integer>();
  for (int i = 0; i < big.length() - small.length() + 1; i++) {
    if (isSubstringAtLocation(big, small, i)) {
      locations.add(i);
    }
  }
  return locations;
}

public static HashMapList<String, Integer> searchAll(String big, String[] smalls) {
  HashMapList<String, Integer> lookup = new HashMapList<String, Integer>();
  for (String small : smalls) {
    ArrayList<Integer> locations = search(big, small);
    lookup.put(small, locations);
  }
  return lookup;
}

AC自动机算法是解决字符串多模式匹配的一个经典方法,时间复杂度为:O(m+kn+z), 其中:m是目标串S的长度,k是模式串个数,n是模式串平均长度,z是S 中出现的模式串数量。从时间复杂度上可以看出,AC自动机比后缀Trie方法要快, m从2次方降到了1次方。
AC自动机也会先构造一棵Trie树,不同的是,它用模式串来构造Trie树。 然后遍历一次目标串S,即可求出哪些模式串出现在目标串S中。
关于AC自动机,比较好的资料是: Set Matching and Aho-Corasick Algorithm 它是 生物序列算法课的一个课件, 这个课的课件基本上都是关于字符串算法的,讲得挺好,推荐一读。


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