实现树形数据结构以查找和搜索最大的人

Question

I've encountered this problem:

some people (with specific id ) want to stand in front of the mirror and no one goes out of the way: the problem gives us the id and height of each person:

 200 4000
 50 1200

which the first number indicates the id and the second one is the height (yes these people are huge.) you can see yourself in the mirror only if you are taller than people in front of you. The order of standing is the same as the order of inputs. after some amount of inputs the problem request whether or not some person with some id can see himself or not? there are two commands: one for adding to the line**+** and one for checking that id**:** so: Input: 100 1000 50 1000 100 50 Output: YES NO

100 can see himself but 50 can not since someone with same height is standing in front of him also if n is the number of command:
also:
no one can change its place and at any time in the n input line the problem can request if a person can see it self or not using this command:

 person's id

The time limit on the problem is one second so I believe we need an O(nlogn) answer.
My solution so far: if we could have an tree with each node not only containing the id and the height but the max height of every body in its sub-tree the it would be very easy to find the answer because we add each node to the tree based on the persons height. I believe red and black trees would be a good data structure here. so for instance I can use TreeSet in java which uses red-black trees.

My problems:

First of all I'm not really sure that my way of thinking is right and what date structure should I use?
second How should I implement this solution (if it's right) in code? The language is not very important but it would be great if you could help using java or c++. I need some hint on how to actually do this algorithm in fastest way

thanks very much

Answer 1

Actually, with these requirements, this can be solved in O(1).

The algorithm:

We keep a heightNeededToSeeSelf variable.

• Person 1 (height = h[0]) always sees themselves as there is no one in front of them.
• neededHeightToSeeSelf = h[0]
• Person 2 (height = h[1]) can see herself if h[1] > neededHeightToSeeSelf .

• if h[1] > neededHeightToSeeSelf :

  neededHeightToSeeSelf = h[1]

• Store a pointer to the id of the person in the line and use it to find the person's status in case asked.
• Continue for person 3,4,...n until user quits.

That's it.

Java code:

public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    Map<String, Integer> idmap = new HashMap<>();
    List<Boolean> canSeeSelf = new ArrayList<>();
    int neededHeightToSeeSelf = 0;
    while(sc.hasNextLine()) {
        String[] line = sc.nextLine().split(" ");
        if(line[0].equals("+")) {
            int personHeight = Integer.parseInt(line[2]);
            if(personHeight > neededHeightToSeeSelf) {
                canSeeSelf.add(true);
                neededHeightToSeeSelf = personHeight;
            }
            else {
                canSeeSelf.add(false);
            }
            idmap.put(line[1], canSeeSelf.size() - 1);
        } else if(line[0].equals("?")) {
            int index = idmap.get(line[1]);
            System.out.println(canSeeSelf.get(index));
        } else {
            System.out.println("Quitting...");
            System.exit(1);
        }
    }
}

idmap simply stores a pointer to the index of the person in the line. This code is only to show the O(1) solution and is not a perfect solution. The indirection can be removed so idmap can directly answer the true/false question.

Test input:

+ 100 1000
+ 50 1000
+ 101 500
+ 102 100
? 102
false
? 100
true
? 50
false
+ 103 2000
? 103
true
end
Quitting...

Answer 2

Can height of the person be updated? Assuming, update of height is not required and the input is given in the order of standing, as you read, check if person height is higher than all other persons added so far(ie currentMaxHeight).

Below is one of the possible solution using Hashmap of Person and Boolean value. This is just a sample code to demonstrate and test.

Person class with id and height, please take closer look at the equals() and hashCode(). public class Person {

private int id;
private int height;

public Person(int id, int height) {
    this.id = id;
    this.height = height;
}

public Person(int id) {
    this.id = id;
}

public int getId() {
    return id;
}

public int getHeight() {
    return height;
}

@Override
public boolean equals(Object o) {
    if (this == o) return true;
    if (o == null || getClass() != o.getClass()) return false;
    Person person = (Person) o;
    return id == person.id;
}

@Override
public int hashCode() {
    return Objects.hash(id);
}

@Override
public String toString() {
    return "Person{" +
            "id=" + id +
            ", height=" + height +
            '}';
}
}

TallestPerson class to add person and verify can see self.

  public class TallestPerson {

    public static void main(String[] args) {

    List<Person> personList = new ArrayList<>();
    personList.add(new Person(100, 100));
    personList.add(new Person(101, 50));
    personList.add(new Person(102, 200));
    personList.add(new Person(103, 200));
    personList.add(new Person(104, 100));
    personList.add(new Person(105, 300));

    addPerson(personList);
    System.out.println(canSeeSelf(100));
    System.out.println(canSeeSelf(101));
    System.out.println(canSeeSelf(102));
    System.out.println(canSeeSelf(103));
    System.out.println(canSeeSelf(104));
    System.out.println(canSeeSelf(105));
}

static int currentMaxHeight = Integer.MIN_VALUE;
static Map<Person, Boolean> personMap = new HashMap<>();

static public void addPerson(List<Person> personList) {
    for (Person person : personList) {
        addPerson(person);
    }
}

static public void addPerson(Person person) {
    addPerson(person.getId(), person.getHeight());
}

static public void addPerson(int id, int height) {
    if (height > currentMaxHeight) {
        personMap.put(new Person(id, height), true);
        currentMaxHeight = height;
    } else {
        personMap.put(new Person(id, height), false);
    }
}

static public boolean canSeeSelf(int id) {
    return personMap.get(new Person(id));
}
}

Output

 true
 false
 true
 false
 false
 true