Leetcode #1600: Throne Inheritance
In this guide, we solve Leetcode #1600 Throne Inheritance in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Depth-First Search, Design, Hash Table
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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
Python Solution
class ThroneInheritance:
def __init__(self, kingName: str):
self.king = kingName
self.dead = set()
self.g = defaultdict(list)
def birth(self, parentName: str, childName: str) -> None:
self.g[parentName].append(childName)
def death(self, name: str) -> None:
self.dead.add(name)
def getInheritanceOrder(self) -> List[str]:
def dfs(x: str):
x not in self.dead and ans.append(x)
for y in self.g[x]:
dfs(y)
ans = []
dfs(self.king)
return ans
# Your ThroneInheritance object will be instantiated and called as such:
# obj = ThroneInheritance(kingName)
# obj.birth(parentName,childName)
# obj.death(name)
# param_3 = obj.getInheritanceOrder()
Complexity
The time complexity is O(n). The space complexity is , where is the number of nodes.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.