1600. Throne Inheritance LeetCode Solution

In this guide, you will get 1600. Throne Inheritance LeetCode Solution with the best time and space complexity. The solution to Throne Inheritance problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Throne Inheritance solution in C++
Throne Inheritance solution in Java
Throne Inheritance solution in Python
Additional Resources

1600. Throne Inheritance LeetCode Solution image

Problem Statement of Throne Inheritance

A kingdom consists of a king, his children, his grandchildren, and so on. Every once in a while, someone in the family dies or a child is born.
The kingdom has a well-defined order of inheritance that consists of the king as the first member. Let’s define the recursive function Successor(x, curOrder), which given a person x and the inheritance order so far, returns who should be the next person after x in the order of inheritance.

Successor(x, curOrder):
if x has no children or all of x’s children are in curOrder:
if x is the king return null
else return Successor(x’s parent, curOrder)
else return x’s oldest child who’s not in curOrder

For example, assume we have a kingdom that consists of the king, his children Alice and Bob (Alice is older than Bob), and finally Alice’s son Jack.

In the beginning, curOrder will be [“king”].
Calling Successor(king, curOrder) will return Alice, so we append to curOrder to get [“king”, “Alice”].
Calling Successor(Alice, curOrder) will return Jack, so we append to curOrder to get [“king”, “Alice”, “Jack”].
Calling Successor(Jack, curOrder) will return Bob, so we append to curOrder to get [“king”, “Alice”, “Jack”, “Bob”].
Calling Successor(Bob, curOrder) will return null. Thus the order of inheritance will be [“king”, “Alice”, “Jack”, “Bob”].

Using the above function, we can always obtain a unique order of inheritance.
Implement the ThroneInheritance class:

ThroneInheritance(string kingName) Initializes an object of the ThroneInheritance class. The name of the king is given as part of the constructor.
void birth(string parentName, string childName) Indicates that parentName gave birth to childName.
void death(string name) Indicates the death of name. The death of the person doesn’t affect the Successor function nor the current inheritance order. You can treat it as just marking the person as dead.
string[] getInheritanceOrder() Returns a list representing the current order of inheritance excluding dead people.

Example 1:

Input
[“ThroneInheritance”, “birth”, “birth”, “birth”, “birth”, “birth”, “birth”, “getInheritanceOrder”, “death”, “getInheritanceOrder”]
[[“king”], [“king”, “andy”], [“king”, “bob”], [“king”, “catherine”], [“andy”, “matthew”], [“bob”, “alex”], [“bob”, “asha”], [null], [“bob”], [null]]
Output
[null, null, null, null, null, null, null, [“king”, “andy”, “matthew”, “bob”, “alex”, “asha”, “catherine”], null, [“king”, “andy”, “matthew”, “alex”, “asha”, “catherine”]]

Explanation
ThroneInheritance t= new ThroneInheritance(“king”); // order: king
t.birth(“king”, “andy”); // order: king > andy
t.birth(“king”, “bob”); // order: king > andy > bob
t.birth(“king”, “catherine”); // order: king > andy > bob > catherine
t.birth(“andy”, “matthew”); // order: king > andy > matthew > bob > catherine
t.birth(“bob”, “alex”); // order: king > andy > matthew > bob > alex > catherine
t.birth(“bob”, “asha”); // order: king > andy > matthew > bob > alex > asha > catherine
t.getInheritanceOrder(); // return [“king”, “andy”, “matthew”, “bob”, “alex”, “asha”, “catherine”]
t.death(“bob”); // order: king > andy > matthew > bob > alex > asha > catherine
t.getInheritanceOrder(); // return [“king”, “andy”, “matthew”, “alex”, “asha”, “catherine”]

Constraints:

1 <= kingName.length, parentName.length, childName.length, name.length <= 15
kingName, parentName, childName, and name consist of lowercase English letters only.
All arguments childName and kingName are distinct.
All name arguments of death will be passed to either the constructor or as childName to birth first.
For each call to birth(parentName, childName), it is guaranteed that parentName is alive.
At most 105 calls will be made to birth and death.
At most 10 calls will be made to getInheritanceOrder.

Complexity Analysis

Time Complexity:
Space Complexity:

1600. Throne Inheritance LeetCode Solution in C++

class ThroneInheritance {
 public:
  ThroneInheritance(string kingName) : kingName(kingName) {}

  void birth(string parentName, string childName) {
    family[parentName].push_back(childName);
  }

  void death(string name) {
    dead.insert(name);
  }

  vector<string> getInheritanceOrder() {
    vector<string> ans;
    dfs(kingName, ans);
    return ans;
  }

 private:
  unordered_set<string> dead;
  unordered_map<string, vector<string>> family;
  string kingName;

  void dfs(const string& name, vector<string>& ans) {
    if (!dead.contains(name))
      ans.push_back(name);
    if (!family.contains(name))
      return;

    for (const string& child : family[name])
      dfs(child, ans);
  }
};
/* code provided by PROGIEZ */

1600. Throne Inheritance LeetCode Solution in Java

class ThroneInheritance {
  public ThroneInheritance(String kingName) {
    this.kingName = kingName;
  }

  public void birth(String parentName, String childName) {
    family.putIfAbsent(parentName, new ArrayList<>());
    family.get(parentName).add(childName);
  }

  public void death(String name) {
    dead.add(name);
  }

  public List<String> getInheritanceOrder() {
    List<String> ans = new ArrayList<>();
    dfs(kingName, ans);
    return ans;
  }

  private Set<String> dead = new HashSet<>();
  private Map<String, List<String>> family = new HashMap<>();
  private String kingName;

  private void dfs(final String name, List<String> ans) {
    if (!dead.contains(name))
      ans.add(name);
    if (!family.containsKey(name))
      return;

    for (final String child : family.get(name))
      dfs(child, ans);
  }
}
// code provided by PROGIEZ