Leetcode #2003: Smallest Missing Genetic Value in Each Subtree
In this guide, we solve Leetcode #2003 Smallest Missing Genetic Value in Each Subtree 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
There is a family tree rooted at 0 consisting of n nodes numbered 0 to n - 1. You are given a 0-indexed integer array parents, where parents[i] is the parent for node i.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Tree, Depth-First Search, Union Find, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: parents = [-1,0,0,2], nums = [1,2,3,4]
Output: [5,1,1,1]
Explanation: The answer for each subtree is calculated as follows:
- 0: The subtree contains nodes [0,1,2,3] with values [1,2,3,4]. 5 is the smallest missing value.
- 1: The subtree contains only node 1 with value 2. 1 is the smallest missing value.
- 2: The subtree contains nodes [2,3] with values [3,4]. 1 is the smallest missing value.
- 3: The subtree contains only node 3 with value 4. 1 is the smallest missing value.
Python Solution
class Solution:
def smallestMissingValueSubtree(
self, parents: List[int], nums: List[int]
) -> List[int]:
def dfs(i: int):
if vis[i]:
return
vis[i] = True
if nums[i] < len(has):
has[nums[i]] = True
for j in g[i]:
dfs(j)
n = len(nums)
ans = [1] * n
g = [[] for _ in range(n)]
idx = -1
for i, p in enumerate(parents):
if i:
g[p].append(i)
if nums[i] == 1:
idx = i
if idx == -1:
return ans
vis = [False] * n
has = [False] * (n + 2)
i = 2
while idx != -1:
dfs(idx)
while has[i]:
i += 1
ans[idx] = i
idx = parents[idx]
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.