Leetcode #339: Nested List Weight Sum
In this guide, we solve Leetcode #339 Nested List Weight Sum 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
You are given a nested list of integers nestedList. Each element is either an integer or a list whose elements may also be integers or other lists.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Depth-First Search, Breadth-First Search
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: nestedList = [[1,1],2,[1,1]]
Output: 10
Explanation: Four 1's at depth 2, one 2 at depth 1. 1*2 + 1*2 + 2*1 + 1*2 + 1*2 = 10.
Python Solution
# """
# This is the interface that allows for creating nested lists.
# You should not implement it, or speculate about its implementation
# """
# class NestedInteger:
# def __init__(self, value=None):
# """
# If value is not specified, initializes an empty list.
# Otherwise initializes a single integer equal to value.
# """
#
# def isInteger(self):
# """
# @return True if this NestedInteger holds a single integer, rather than a nested list.
# :rtype bool
# """
#
# def add(self, elem):
# """
# Set this NestedInteger to hold a nested list and adds a nested integer elem to it.
# :rtype void
# """
#
# def setInteger(self, value):
# """
# Set this NestedInteger to hold a single integer equal to value.
# :rtype void
# """
#
# def getInteger(self):
# """
# @return the single integer that this NestedInteger holds, if it holds a single integer
# Return None if this NestedInteger holds a nested list
# :rtype int
# """
#
# def getList(self):
# """
# @return the nested list that this NestedInteger holds, if it holds a nested list
# Return None if this NestedInteger holds a single integer
# :rtype List[NestedInteger]
# """
class Solution:
def depthSum(self, nestedList: List[NestedInteger]) -> int:
def dfs(nestedList, depth):
depth_sum = 0
for item in nestedList:
if item.isInteger():
depth_sum += item.getInteger() * depth
else:
depth_sum += dfs(item.getList(), depth + 1)
return depth_sum
return dfs(nestedList, 1)
Complexity
The time complexity is O(V+E). The space complexity is O(V).
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.