## Wednesday, May 16, 2018

### LeetCode 745 - Prefix and Suffix Search

https://leetcode.com/problems/prefix-and-suffix-search/description/
Given many wordswords[i] has weight i.
Design a class WordFilter that supports one function, WordFilter.f(String prefix, String suffix). It will return the word with given prefix and suffix with maximum weight. If no word exists, return -1.
Examples:
Input:
WordFilter(["apple"])
WordFilter.f("a", "e") // returns 0
WordFilter.f("b", "") // returns -1

Note:
1. words has length in range [1, 15000].
2. For each test case, up to words.length queries WordFilter.f may be made.
3. words[i] has length in range [1, 10].
4. prefix, suffix have lengths in range [0, 10].
5. words[i] and prefix, suffix queries consist of lowercase letters only.

Consider the word 'apple'. For each suffix of the word, we could insert that suffix, followed by '#', followed by the word, all into the trie.
For example, we will insert '#apple', 'e#apple', 'le#apple', 'ple#apple', 'pple#apple', 'apple#apple' into the trie. Then for a query like prefix = "ap", suffix = "le", we can find it by querying our trie for le#ap.
• Time Complexity: $O(NK^2 + QK)$ where $N$ is the number of words, $K$ is the maximum length of a word, and $Q$ is the number of queries.
• Space Complexity: $O(NK^2)$, the size of the trie.
class WordFilter {
TrieNode trie;
public WordFilter(String[] words) {
trie = new TrieNode();
for (int weight = 0; weight < words.length; ++weight) {
String word = words[weight] + "{";
for (int i = 0; i < word.length(); ++i) {
TrieNode cur = trie;
cur.weight = weight;
for (int j = i; j < 2 * word.length() - 1; ++j) {
int k = word.charAt(j % word.length()) - 'a';
if (cur.children[k] == null)
cur.children[k] = new TrieNode();
cur = cur.children[k];
cur.weight = weight;
}
}
}
}
public int f(String prefix, String suffix) {
TrieNode cur = trie;
for (char letter: (suffix + '{' + prefix).toCharArray()) {
if (cur.children[letter - 'a'] == null) return -1;
cur = cur.children[letter - 'a'];
}
return cur.weight;
}
}

class TrieNode {
TrieNode[] children;
int weight;
public TrieNode() {
children = new TrieNode[27];
weight = 0;
}
}

Say we are inserting the word apple. We could insert ('a', 'e'), ('p', 'l'), ('p', 'p'), ('l', 'p'), ('e', 'a') into our trie. Then, if we had equal length queries like prefix = "ap", suffix = "le", we could find the node trie['a', 'e']['p', 'l'] in our trie. This seems promising.
What about queries that aren't equal? We should just insert them like normal. For example, to capture a case like prefix = "app", suffix = "e", we could create nodes trie['a', 'e']['p', None]['p', None].
After inserting these pairs into our trie, our searches are straightforward.
• Time Complexity: $O(NK^2 + QK)$ where $N$ is the number of words, $K$ is the maximum length of a word, and $Q$ is the number of queries.
• Space Complexity: $O(NK^2)$, the size of the trie.
class WordFilter {
TrieNode trie;
public WordFilter(String[] words) {
trie = new TrieNode();
int wt = 0;
for (String word: words) {
TrieNode cur = trie;
cur.weight = wt;
int L = word.length();
char[] chars = word.toCharArray();
for (int i = 0; i < L; ++i) {

TrieNode tmp = cur;
for (int j = i; j < L; ++j) {
int code = (chars[j] - '') * 27;
if (tmp.children.get(code) == null)
tmp.children.put(code, new TrieNode());
tmp = tmp.children.get(code);
tmp.weight = wt;
}

tmp = cur;
for (int k = L - 1 - i; k >= 0; --k) {
int code = (chars[k] - '');
if (tmp.children.get(code) == null)
tmp.children.put(code, new TrieNode());
tmp = tmp.children.get(code);
tmp.weight = wt;
}

int code = (chars[i] - '') * 27 + (chars[L - 1 - i] - '');
if (cur.children.get(code) == null)
cur.children.put(code, new TrieNode());
cur = cur.children.get(code);
cur.weight = wt;

}
wt++;
}
}

public int f(String prefix, String suffix) {
TrieNode cur = trie;
int i = 0, j = suffix.length() - 1;
while (i < prefix.length() || j >= 0) {
char c1 = i < prefix.length() ? prefix.charAt(i) : '';
char c2 = j >= 0 ? suffix.charAt(j) : '';
int code = (c1 - '') * 27 + (c2 - '');
cur = cur.children.get(code);
if (cur == null) return -1;
i++; j--;
}
return cur.weight;
}
}

class TrieNode {
Map<Integer, TrieNode> children;
int weight;
public TrieNode() {
children = new HashMap();
weight = 0;
}
}

We use two tries to separately find all words that match the prefix, plus all words that match the suffix. Then, we try to find the highest weight element in the intersection of these sets.
Of course, these sets could still be large, so we might TLE if we aren't careful.

• Time Complexity: $O(NK + Q(N+K))$ where $N$ is the number of words, $K$ is the maximum length of a word, and $Q$ is the number of queries. If we use memoization in our solution, we could produce tighter bounds for this complexity, as the complex queries are somewhat disjoint.
• Space Complexity: $O(NK)$, the size of the tries.
class WordFilter {
TrieNode trie1, trie2;
public WordFilter(String[] words) {
trie1 = new TrieNode();
trie2 = new TrieNode();
int wt = 0;
for (String word: words) {
char[] ca = word.toCharArray();

TrieNode cur = trie1;
for (char letter: ca) {
if (cur.children[letter - 'a'] == null)
cur.children[letter - 'a'] = new TrieNode();
cur = cur.children[letter - 'a'];
}

cur = trie2;
for (int j = ca.length - 1; j >= 0; --j) {
char letter = ca[j];
if (cur.children[letter - 'a'] == null)
cur.children[letter - 'a'] = new TrieNode();
cur = cur.children[letter - 'a'];
}
wt++;
}
}

public int f(String prefix, String suffix) {
TrieNode cur1 = trie1, cur2 = trie2;
for (char letter: prefix.toCharArray()) {
if (cur1.children[letter - 'a'] == null) return -1;
cur1 = cur1.children[letter - 'a'];
}
char[] ca = suffix.toCharArray();
for (int j = ca.length - 1; j >= 0; --j) {
char letter = ca[j];
if (cur2.children[letter - 'a'] == null) return -1;
cur2 = cur2.children[letter - 'a'];
}

int ans = -1;
for (int w1: cur1.weight)
if (w1 > ans && cur2.weight.contains(w1))
ans = w1;

return ans;
}
}

class TrieNode {
TrieNode[] children;
Set<Integer> weight;
public TrieNode() {
children = new TrieNode[26];
weight = new HashSet();
}
}

https://leetcode.com/problems/prefix-and-suffix-search/discuss/110044/Three-ways-to-solve-this-problem-in-Java
If f is more frequently than WordFilter, use method 1.
If space complexity is concerned, use method 2.
If the input string array might change frequently, use method 3.
< Method 1 >
WordFilter: Time = O(NL^2)
f: Time = O(1)
Space = O(NL^2)
Note: N is the size of input array and L is the max length of input strings.
class WordFilter {
HashMap<String, Integer> map = new HashMap<>();

public WordFilter(String[] words) {
for(int w = 0; w < words.length; w++){
for(int i = 0; i <= 10 && i <= words[w].length(); i++){
for(int j = 0; j <= 10 && j <= words[w].length(); j++){
map.put(words[w].substring(0, i) + "#" + words[w].substring(words[w].length()-j), w);
}
}
}
}

public int f(String prefix, String suffix) {
return (map.containsKey(prefix + "#" + suffix))? map.get(prefix + "#" + suffix) : -1;
}
}


< Method 2 >
WordFilter: Time = O(NL)
f: Time = O(N)
Space = O(NL)
class WordFilter {
HashMap<String, List<Integer>> mapPrefix = new HashMap<>();
HashMap<String, List<Integer>> mapSuffix = new HashMap<>();

public WordFilter(String[] words) {

for(int w = 0; w < words.length; w++){
for(int i = 0; i <= 10 && i <= words[w].length(); i++){
String s = words[w].substring(0, i);
if(!mapPrefix.containsKey(s)) mapPrefix.put(s, new ArrayList<>());
}
for(int i = 0; i <= 10 && i <= words[w].length(); i++){
String s = words[w].substring(words[w].length() - i);
if(!mapSuffix.containsKey(s)) mapSuffix.put(s, new ArrayList<>());
}
}

public int f(String prefix, String suffix) {

if(!mapPrefix.containsKey(prefix) || !mapSuffix.containsKey(suffix)) return -1;
List<Integer> p = mapPrefix.get(prefix);
List<Integer> s = mapSuffix.get(suffix);
int i = p.size()-1, j = s.size()-1;
while(i >= 0 && j >= 0){
if(p.get(i) < s.get(j)) j--;
else if(p.get(i) > s.get(j)) i--;
else return p.get(i);
}
return -1;

< Method 3 >
WordFilter: Time = O(1)
f: Time = O(NL)
Space = O(1)
class WordFilter {
String[] input;
public WordFilter(String[] words) {
input = words;
}
public int f(String prefix, String suffix) {
for(int i = input.length-1; i >= 0; i--){
if(input[i].startsWith(prefix) && input[i].endsWith(suffix)) return i;
}
return -1;
}
}
`