Problem Description

In this problem, we are given a nestedList consisting of integers that might be wrapped in multiple layers of lists. An integer's depth is determined by how many lists contain it. For instance, in the list [1,[2,2],[[3],2],1], the number 1 is at depth 1 (since it's not within any nested list), the number 2 is at depth 2 (as it is in one nested list), and the number 3 is at depth 3 (as it's in two nested lists).

Our goal is to compute the sum of all integers in nestedList each weighted by its depth. In simple terms, we multiply each integer by the number of lists it is inside and then sum these products together.

Flowchart Walkthrough

To analyze the problem defined in Leetcode 339. Nested List Weight Sum, using the Flowchart, let's follow the steps to identify the applicable algorithm:

Is it a graph?

• Yes: Although not explicitly stated as a traditional graph, the nested structure can be visualized as a graph where each nested list is a node, and each element or nested sublist within is a connected node.

Is it a tree?

• Yes: The data structure is recursively nested, which can be mirrored as a tree with children being either elements or further nested lists.

Following the tree path in the flowchart, our next algorithmic approach points directly to using:

• DFS: Depth-First Search is well-suited for hierarchically structured data or problems where you need to explore the depths of each branch, such as calculating weights of nested lists where each level has a multiplicative factor depending on its depth.

Conclusion: The flowchart suggests using DFS for efficiently calculating the weighted sum of each element in nested lists in a tree-like data structure.

Intuition

To arrive at the solution, we can utilize a depth-first search (DFS) strategy, where we traverse the nested list structure recursively. Whenever we encounter an integer, we multiply it by its depth and add it to the cumulative sum. If we encounter a nested list within this list, we call the DFS function recursively with the nested list and an incremented depth value.

The intuition behind using DFS is that it allows us to handle an arbitrary level of nesting in the list by recursively processing nested lists until we reach the most deeply nested integers. This method naturally handles the nested structure and allows us to apply the depth multiplier as we traverse the data.

Learn more about Depth-First Search and Breadth-First Search patterns.

Solution Approach

The solution uses a recursive function, dfs, which is a common algorithm in dealing with tree or graph data structures. In this case, the nested lists can be imagined as a tree where each list is a node, and its elements are either child nodes (other lists) or leaf nodes (integers).

The dfs function has two parameters, nestedList, which is the current list to be processed, and depth, which represents the current depth of traversal.

Here is a step-by-step breakdown of the dfs function:

1. Initialize a local variable depth_sum to 0. This variable will hold the sum of the integers multiplied by their respective depths for the current list.

2. Iterate through each item in the nestedList:

   • If the item is an integer (checked using item.isInteger()), multiply the integer by its depth (item.getInteger() * depth) and add the result to depth_sum.
   • If the item is a nested list, make a recursive call to dfs with item.getList() and depth + 1 to handle the deeper level, and add the result to depth_sum.

3. After iterating through all items, return depth_sum to the caller.

The function kicks off by calling dfs with the initial nestedList and a starting depth of 1. As dfs progresses through each level of nesting, it adds to the cumulative sum while correctly adjusting the depth.

Effectively, this approach applies a depth-first traversal on the tree structure of nested lists, calculating the product of the integer values and their respective depths at each node and aggregating these values up the call stack.

Here is the main function call:

return dfs(nestedList, 1)

It starts the recursive process and eventually returns the required weighted sum after the entire nested structure has been explored.

Example Walkthrough

Let's explain the depth-first search (DFS) approach with a small example, consider the nested list [1, [2, 2], [[3], 2], 1].

1. The DFS process starts by calling dfs(nestedList, 1) with the initial depth as 1.
2. Starting the DFS algorithm, the function initializes a local variable depth_sum to 0.
3. The function begins to iterate over each item in the nestedList.
4. The first element 1 is an integer and not a nested list.
   • Since it’s at depth 1, we multiply it by its depth: 1 * 1.
   • We add this to depth_sum, which is 0 + 1 = 1.
5. The second element [2, 2] is a nested list, so we call dfs([2, 2], 2) because we are now one layer deeper.
   • This call itself will add 2 * 2 + 2 * 2 to our cumulative sum because both 2s are at depth 2.
   • After this recursive call, depth_sum is updated by 1 + 8 = 9.
6. The third element [[3], 2] is another nested list. We now encounter two cases:
   • The nested list [3] requires a recursive call: dfs([3], 3).
     • The element 3 at depth 3 gives 3 * 3, and the depth_sum is increased by 9.
   • The integer 2 is at depth 2, resulting in 2 * 2.
   • Combining these, we add 9 + 4 = 13 to our depth_sum, resulting in 9 + 13 = 22.
7. Finally, the last element 1 is an integer.
   • It’s at depth 1, so it contributes 1 * 1 to our sum.
   • The final depth_sum is updated to 22 + 1 = 23.

Therefore, the result of the DFS algorithm when applied to our example nested list [1, [2, 2], [[3], 2], 1] yields 23. This is how the recursive function cumulatively builds up the sum of the products of integers and their respective depths as it iterates through the nested list.

Solution Implementation

Time and Space Complexity

The given code defines a function dfs that traverses a nested list structure to compute the weighted sum of all the integers within, where each integer is multiplied by its depth level.

Time Complexity

The time complexity of the function is O(N), where N is the total number of integers and lists within all levels of the nested list. Specifically, the function dfs visits each element exactly once. For each integer it encounters, it performs a constant time operation of multiplication and addition. For each list, it makes a recursive call to process its elements. However, since every element is only visited once, the overall time to visit all elements is proportional to their count.

Space Complexity

The space complexity of the function is O(D), where D is the maximum depth of the nested list. This complexity arises from the call stack used for recursion. In the worst case, the recursion will go as deep as the deepest nested list. Therefore, the maximum number of nested calls will equal the maximum depth D. Furthermore, there is only a constant amount of space used at each level for variables such as depth_sum.

Learn more about how to find time and space complexity quickly using problem constraints.