LeetCode #1600 — MEDIUM

Throne Inheritance

Move from brute-force thinking to an efficient approach using hash map strategy.

Solve on LeetCode
The Problem

Problem Statement

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.

  1. In the beginning, curOrder will be ["king"].
  2. Calling Successor(king, curOrder) will return Alice, so we append to curOrder to get ["king", "Alice"].
  3. Calling Successor(Alice, curOrder) will return Jack, so we append to curOrder to get ["king", "Alice", "Jack"].
  4. Calling Successor(Jack, curOrder) will return Bob, so we append to curOrder to get ["king", "Alice", "Jack", "Bob"].
  5. 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.
Patterns Used

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Tree · Design

Example 1

["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]]

Related Problems

  • Operations on Tree (operations-on-tree)
Step 02

Core Insight

What unlocks the optimal approach

  • Create a tree structure of the family.
  • Without deaths, the order of inheritance is simply a pre-order traversal of the tree.
  • Mark the dead family members tree nodes and don't include them in the final order.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1600: Throne Inheritance
class ThroneInheritance {
    private String king;
    private Set<String> dead = new HashSet<>();
    private Map<String, List<String>> g = new HashMap<>();
    private List<String> ans = new ArrayList<>();

    public ThroneInheritance(String kingName) {
        king = kingName;
    }

    public void birth(String parentName, String childName) {
        g.computeIfAbsent(parentName, k -> new ArrayList<>()).add(childName);
    }

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

    public List<String> getInheritanceOrder() {
        ans.clear();
        dfs(king);
        return ans;
    }

    private void dfs(String x) {
        if (!dead.contains(x)) {
            ans.add(x);
        }
        for (String y : g.getOrDefault(x, List.of())) {
            dfs(y);
        }
    }
}

/**
 * Your ThroneInheritance object will be instantiated and called as such:
 * ThroneInheritance obj = new ThroneInheritance(kingName);
 * obj.birth(parentName,childName);
 * obj.death(name);
 * List<String> param_3 = obj.getInheritanceOrder();
 */
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(h)

Approach Breakdown

LEVEL ORDER
O(n) time
O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL
O(n) time
O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.