Trie 3 | Leetcode 1268. Search Suggestions System | Amazon | OA 2020 | Keyword Suggestions

 Question:

Given an array of strings products and a string searchWord. We want to design a system that suggests at most three product names from products after each character of searchWord is typed. Suggested products should have common prefix with the searchWord. If there are more than three products with a common prefix return the three lexicographically minimums products.

Return list of lists of the suggested products after each character of searchWord is typed. 

Example 1:

Input: products = ["mobile","mouse","moneypot","monitor","mousepad"], searchWord = "mouse"

Output: [

["mobile","moneypot","monitor"],

["mobile","moneypot","monitor"],

["mouse","mousepad"],

["mouse","mousepad"],

["mouse","mousepad"]

]

Explanation: products sorted lexicographically = ["mobile","moneypot","monitor","mouse","mousepad"]

After typing m and mo all products match and we show user ["mobile","moneypot","monitor"]

After typing mou, mous and mouse the system suggests ["mouse","mousepad"]

Example 2:

Input: products = ["havana"], searchWord = "havana"

Output: [["havana"],["havana"],["havana"],["havana"],["havana"],["havana"]]

Example 3:

Input: products = ["bags","baggage","banner","box","cloths"], searchWord = "bags"

Output: [["baggage","bags","banner"],["baggage","bags","banner"],["baggage","bags"],["bags"]]

Example 4:

Input: products = ["havana"], searchWord = "tatiana"

Output: [[],[],[],[],[],[],[]]

Solution :

class Solution {

    class TrieNode{

        TrieNode[] children=new TrieNode[26];

        LinkedList<String> suggestions=new LinkedList<>();

        boolean isEnd=false;

    }

    

    private void insert(String p,TrieNode root){

        TrieNode node=root;

        for(char ch:p.toCharArray()){

            if(node.children[ch-'a']==null){

               node.children[ch-'a']=new TrieNode(); 

            }

            node=node.children[ch-'a'];

            //add the word in the suggestions

            node.suggestions.add(p);

            Collections.sort(node.suggestions);

            

            if(node.suggestions.size() >3){

               node.suggestions.pollLast(); 

            }

        }

    }

    private List<List<String>> search(String searchWord,TrieNode root){

        List<List<String>> ans=new ArrayList<>();

        TrieNode node=root;

        for(char ch:searchWord.toCharArray()){

            if(node!=null)

                node=node.children[ch-'a'];

            ans.add(node== null ? Arrays.asList():node.suggestions);

            

        }

        return ans;

        

    }

    

    public List<List<String>> suggestedProducts(String[] products, String searchWord) {

        TrieNode root=new TrieNode();

        for(String s:products){

            insert(s,root);

        }

        return search(searchWord,root);

    }

}

Analysis:

Complexity depends on the sorting, the process of building Trie and the length of searchWord. Sorting cost time O(m * n), due to involving comparing String, which cost time O(m) for each comparison, building Trie cost O(m * n). Therefore,

Time: O(m * n + L), space: O(m * n + L * m) - including return list ans, where m = average length of products, n = products.length, L = searchWord.length().

Trie | Part 2 Implement Trie (Prefix Tree)

Leetcode 208. Implement Trie (Prefix Tree)

Implement a trie with insert, search, and startsWith methods.

Example:

Trie trie = new Trie();

trie.insert("apple");

trie.search("apple");   // returns true

trie.search("app");     // returns false

trie.startsWith("app"); // returns true

trie.insert("app");   

trie.search("app");     // returns true

Note:

You may assume that all inputs are consist of lowercase letters a-z.

All inputs are guaranteed to be non-empty strings.

Solution :

class Trie {

    

    static class TrieNode{

        TrieNode[] children=new TrieNode[26];

        boolean isEnd=false;

    }

    

    private TrieNode root;

    /** Initialize your data structure here. */

    public Trie() {

        root=new TrieNode();

    }

    

    /** Inserts a word into the trie. */

    public void insert(String word) {

        TrieNode node=root;

        for(int i=0;i<word.length();i++){

            char ch=word.charAt(i);

            if(node.children[ch-'a']==null){

                node.children[ch-'a']=new TrieNode();

            }

            node=node.children[ch-'a'];

        }

        node.isEnd=true;

        

    }

    

    /** Returns if the word is in the trie. */

    public boolean search(String word) {

        TrieNode node=root;

        for(int i=0;i<word.length();i++){

            char ch=word.charAt(i);

            if(node.children[ch-'a']==null)

                return false;

            node=node.children[ch-'a'];

        }

        return node.isEnd;

        

    }

    

    /** Returns if there is any word in the trie that starts with the given prefix. */

    public boolean startsWith(String prefix) {

        TrieNode node=root;

        for(int i=0;i<prefix.length();i++){

            char ch=prefix.charAt(i);

            if(node.children[ch-'a']==null)

                return false;

            node=node.children[ch-'a'];

        }

        return node != null;

    }

}

/**

 * Your Trie object will be instantiated and called as such:

 * Trie obj = new Trie();

 * obj.insert(word);

 * boolean param_2 = obj.search(word);

 * boolean param_3 = obj.startsWith(prefix);

 */

Trie | Part 1 Introduction

 Trie (pronounce as "Try") is a very popular data structure, usually known as Prefix Tree as well.

Applications:

1. AutoComplete

2. Spell Checker

3. IP Routing (Longest Prefix Matching)

There are several other data structures, like balanced trees and hash tables, which give us the possibility to search for a word in a dataset of strings. Then why do we need trie? Although hash table has O(1) time complexity for looking for a key, it is not efficient in the following operations :

Finding all keys with a common prefix.

Enumerating a dataset of strings in lexicographical order.

Another reason why trie outperforms hash table, is that as hash table increases in size, there are lots of hash collisions and the search time complexity could deteriorate to O(n), where n is the number of keys inserted. Trie could use less space compared to Hash Table when storing many keys with the same prefix. In this case using trie has only O(m) time complexity, where mm is the key length. Searching for a key in a balanced tree costs O(mlogn) time complexity.

Trie node structure

Trie is a rooted tree. Its nodes have the following fields:

Maximum of R links to its children, where each link corresponds to one of R character values from dataset alphabet. In this article we assume that R is 26, the number of lowercase latin letters.

Boolean field which specifies whether the node corresponds to the end of the key, or is just a key prefix.

static class TrieNode{

        TrieNode[] children=new TrieNode[26];
        boolean isEnd=false;
    }

Two of the most common operations in a trie are insertion of a key and search for a key.

Insertion of a key to a trie

We insert a key by searching into the trie. We start from the root and search a link, which corresponds to the first key character. There are two cases :

A link exists. Then we move down the tree following the link to the next child level. The algorithm continues with searching for the next key character.

A link does not exist. Then we create a new node and link it with the parent's link matching the current key character. We repeat this step until we encounter the last character of the key, then we mark the current node as an end node and the algorithm finishes.

    /** Inserts a word into the trie. */

    public void insert(String word) {

        TrieNode node=root;

        for(int i=0;i<word.length();i++){

            char ch=word.charAt(i);

            if(node.children[ch-'a']==null){

                node.children[ch-'a']=new TrieNode();

            }

            node=node.children[ch-'a'];

        }

        node.isEnd=true;

        

    }

Complexity Analysis

Time complexity : O(m), where m is the key length.

In each iteration of the algorithm, we either examine or create a node in the trie till we reach the end of the key. This takes only mm operations.

Space complexity : O(m).

In the worst case newly inserted key doesn't share a prefix with the the keys already inserted in the trie. We have to add mm new nodes, which takes us O(m)O(m) space.

Search for a key in a trie

Each key is represented in the trie as a path from the root to the internal node or leaf. We start from the root with the first key character. We examine the current node for a link corresponding to the key character. There are two cases :

A link exist. We move to the next node in the path following this link, and proceed searching for the next key character.

A link does not exist. If there are no available key characters and current node is marked as isEnd we return true. Otherwise there are possible two cases in each of them we return false :

There are key characters left, but it is impossible to follow the key path in the trie, and the key is missing.

No key characters left, but current node is not marked as isEnd. Therefore the search key is only a prefix of another key in the trie.

  /** Returns if the word is in the trie. */

    public boolean search(String word) {

        TrieNode node=root;

        for(int i=0;i<word.length();i++){

            char ch=word.charAt(i);

            if(node.children[ch-'a']==null)

                return false;

            node=node.children[ch-'a'];

        }

        return node.isEnd;

        

    }

Complexity Analysis

Time complexity : O(m)

Space complexity : O(1)

Reference : https://leetcode.com/problems/implement-trie-prefix-tree/solution/

Deleting data from ArrayList with a For-loop

 Suppose you want to remove all the Strings which contains "abc" in them.

for (int i = 0; i < size; i++){

    if (data.get(i).getCaption().contains("abc")){

        data.remove(i);

    }

}

The above code will not work. Why?

The Problem here is you are iterating from 0 to size and inside the loop you are deleting items. Deleting the items will reduce the size of the list which will fail when you try to access the indexes which are greater than the effective size(the size after the deleted items).

There are three approaches to do this.

Delete using iterator if you do not want to deal with index.

for (Iterator<Object> it = data.iterator(); it.hasNext();) {

if (it.next().getCaption().contains("_Hardi")) {

    it.remove();

}

}

Else, delete from the end.

for (int i = size-1; i >= 0; i--){

    if (data.get(i).getCaption().contains("_Hardi")){

            data.remove(i);

    }

 }

Otherwise use a while loop :

int i = 0;
while (i < data.size()) {
    if (data.get(i).getCaption().contains("_Hardi"))
        data.remove(i);
    else i++;
}

Plus, Keep an eye for ConcurrentModificationExceltion as well, although it 

will not com in this case.