1032. Stream of Characters

Problem

Design an algorithm that accepts a stream of characters and checks if a suffix of these characters is a string of a given array of strings words.

For example, if words = ["abc", "xyz"] and the stream added the four characters (one by one) 'a', 'x', 'y', and 'z', your algorithm should detect that the suffix "xyz" of the characters "axyz" matches "xyz" from words.

Implement the StreamChecker class:

• StreamChecker(String[] words) Initializes the object with the strings array words.

• boolean query(char letter) Accepts a new character from the stream and returns true if any non-empty suffix from the stream forms a word that is in words.

  Example 1:

Input
["StreamChecker", "query", "query", "query", "query", "query", "query", "query", "query", "query", "query", "query", "query"]
[[["cd", "f", "kl"]], ["a"], ["b"], ["c"], ["d"], ["e"], ["f"], ["g"], ["h"], ["i"], ["j"], ["k"], ["l"]]
Output
[null, false, false, false, true, false, true, false, false, false, false, false, true]

Explanation
StreamChecker streamChecker = new StreamChecker(["cd", "f", "kl"]);
streamChecker.query("a"); // return False
streamChecker.query("b"); // return False
streamChecker.query("c"); // return False
streamChecker.query("d"); // return True, because 'cd' is in the wordlist
streamChecker.query("e"); // return False
streamChecker.query("f"); // return True, because 'f' is in the wordlist
streamChecker.query("g"); // return False
streamChecker.query("h"); // return False
streamChecker.query("i"); // return False
streamChecker.query("j"); // return False
streamChecker.query("k"); // return False
streamChecker.query("l"); // return True, because 'kl' is in the wordlist

  Constraints:

• 1 <= words.length <= 2000

• 1 <= words[i].length <= 200

• words[i] consists of lowercase English letters.

• letter is a lowercase English letter.

• At most 4 * 10^4 calls will be made to query.

Solution

class StreamChecker {
    static class Node {
        Node[] child;
        boolean isEnd;

        Node() {
            child = new Node[26];
        }
    }

    private final StringBuilder sb;
    private final Node root;

    public void insert(String s) {
        Node curr = root;
        for (int i = s.length() - 1; i >= 0; i--) {
            char c = s.charAt(i);
            if (curr.child[c - 'a'] == null) {
                curr.child[c - 'a'] = new Node();
            }
            curr = curr.child[c - 'a'];
        }
        curr.isEnd = true;
    }

    public StreamChecker(String[] words) {
        root = new Node();
        sb = new StringBuilder();
        for (String s : words) {
            insert(s);
        }
    }

    public boolean query(char letter) {
        sb.append(letter);
        Node curr = root;
        for (int i = sb.length() - 1; i >= 0; i--) {
            char c = sb.charAt(i);
            if (curr.child[c - 'a'] == null) {
                return false;
            }
            if (curr.child[c - 'a'].isEnd) {
                return true;
            }
            curr = curr.child[c - 'a'];
        }
        return false;
    }
}

/**
 * Your StreamChecker object will be instantiated and called as such:
 * StreamChecker obj = new StreamChecker(words);
 * boolean param_1 = obj.query(letter);
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).