339. Nested List Weight Sum - Detailed Explanation

In this optimized BFS approach, we use the fact that we're processing in fixed-depth "waves." The variable depth increases once per level, rather than being stored per element. This removes the slight redundancy of incrementing and storing depth for every single element. The code is concise and still easy to follow. It’s essentially the same complexity as the standard BFS solution, but it’s a nice demonstration of using level traversal to manage state globally.

Step-by-Step Example Walkthrough

Let’s walk through Example 2 (nestedList = [1,[4,[6]]]) step by step to see how the algorithm (DFS recursive approach) processes it and accumulates the sum:

nestedList = [ 1, [4, [6]] ]

• Start at depth 1: The list has two elements at the top level. Initialize sum = 0.

  • First element is 1 (an integer). Depth = 1, so add 1 * 1 to sum.
    • Current sum = 1.
  • Second element is [4, [6]] (a list). It’s not an integer, so we go one level deeper into this sublist.

• Depth 2 (inside the sublist [4, [6]]): Now we are examining the list [4, [6]] with depth = 2.

  • First element here is 4 (integer). Depth = 2, add 4 * 2 to sum.
    • Current sum = 1 (previous) + 8 = 9.
  • Second element is [6] (a list). Not an integer, go one level deeper into this sublist.

• Depth 3 (inside the sublist [6]): We are now looking at the innermost list [6] at depth = 3.

  • It contains a single element 6 (integer). Depth = 3, add 6 * 3 to sum.
    • Current sum = 9 (previous) + 18 = 27.
  • This sublist has no more elements. We step out to the previous level.

• We have finished processing all elements of the depth-2 list [4, [6]]. Step back out to depth 1 context.

• Back at depth 1, there are no more elements in the top-level list to process.

• Final sum = 27. This matches the expected output.

The above trace demonstrates a DFS (depth-first) process. The algorithm keeps track of the current depth as it goes down into nested lists and back up. In an iterative DFS using a stack, the operations would occur in a similar order (though the exact sequence of pushing/popping might differ slightly).

If we used BFS (level-order) for the same example, the order of processing would be different, but the result would be the same:

• Depth 1: process 1 (add 1*1), encounter sublist [4,[6]] (enqueue its contents for next level).
• Depth 2: process 4 (add 4*2), encounter sublist [6] (enqueue its contents).
• Depth 3: process 6 (add 6*3).
• Sum = 27.

Both approaches correctly accumulate the weighted sum.

Common Mistakes and Edge Cases

• Incorrect Depth Updates: A frequent bug is forgetting to increment the depth when going into a nested list, or using the wrong depth for an integer. Always increment depth as you go deeper, and if using recursion, be careful to not carry over depth incorrectly when backtracking (this is naturally handled if depth is a function parameter or a local variable). In an iterative solution, ensure that when you push a child element, you use depth+1. Forgetting these will lead to wrong weights.
• Mixing up Depth vs. Weight in Variation: If you attempt the inverted weight variation (Nested List Weight Sum II), be cautious not to confuse the normal depth with the “inverted weight.” Many mistakes happen by using the depth directly instead of the maxDepth - depth + 1 formula for weights, or by not updating the logic to accumulate weights properly.
• Not Handling Empty Lists: The nested list might contain empty sublists (e.g., [1, [], [2,[ ]]]). Your code should handle these gracefully (they contribute nothing to the sum and should simply be skipped over). Ensure that when you get a list, you iterate its children even if that list could be empty. Similarly, an entire nestedList could theoretically be empty (though the problem constraints ensure at least 1 element in top list). An empty input should return 0.
• Edge Case - Single Integer: If the input is just a single integer (e.g., [5] with no nesting), the output should just be 5 (depth 1). Make sure your code correctly handles the top-level integers. All approaches above do handle this, but if you wrote something fancy, double-check that a non-nested scenario works.
• Using Global Variables in Recursion: If implementing recursively in a real interview or coding environment, watch out for using global variables for accumulating sum or depth. It’s safer to pass depth as a parameter or use a helper function that returns the sum. Globals can lead to errors if the function is used multiple times in the same program or if not reset properly.
• Stack/Queue Initialization: In iterative approaches, a common mistake is not initializing the stack/queue with all top-level elements, or initializing depth incorrectly. Ensure each top-level element is enqueued with depth 1 (or pushed with depth 1). If you only push the first element, you’ll miss others. Also, be careful with the order if that matters for your traversal logic.

Alternative Variation: Nested List Weight Sum II (Inverse Depth Weighting)

There is a follow-up variation of this problem, often known as Nested List Weight Sum II (LeetCode 364). In this variation, the weight of integers is reversed: instead of deeper integers having a larger weight, they have a smaller weight. Specifically, the weight is defined such that leaf-level (deepest) integers have weight 1, and the weight increases by 1 for each level you go up . Equivalently, if maxDepth is the maximum depth of any integer in the list, an integer at depth d in the original sense now has weight = maxDepth - d + 1 .

To clarify, consider the same examples with this inverted weighting:

• Example 1 (Inverse weights): nestedList = [[1,1],2,[1,1]]. Here maxDepth = 2. The four 1's are at depth 2, so their weight is 2 - 2 + 1 = 1. The integer 2 is at depth 1, so its weight is 2 - 1 + 1 = 2. The sum becomes 1*1 + 1*1 + 1*1 + 1*1 + 2*2 = 8 .
• Example 2 (Inverse weights): nestedList = [1,[4,[6]]]. Here maxDepth = 3. The integer 1 (depth 1) gets weight 3 - 1 + 1 = 3; 4 (depth 2) gets weight 3 - 2 + 1 = 2; 6 (depth 3) gets weight 3 - 3 + 1 = 1. The weighted sum is 1*3 + 4*2 + 6*1 = 17.

In other words, the shallower the integer, the larger its weight in this variation (the opposite of the original problem).

Approach to Solve (Two-Pass): A straightforward way to solve Nested List Weight Sum II is:

1. Find maxDepth: Do a DFS (or BFS) traversal to determine the maximum depth of any integer in the nested list. This is similar to computing the maximum nesting level.
2. Compute Weighted Sum: Traverse the list again, this time using the formula weight = maxDepth - currentDepth + 1 for each integer. Sum up value * weight for all integers.

For example, in Python:

def depthSumInverse_two_pass(nestedList):
    # First pass to find maximum depth
    def get_max_depth(lst, depth):
        max_d = depth
        for ni in lst:
            if not ni.isInteger():
                max_d = max(max_d, get_max_depth(ni.getList(), depth+1))
        return max_d
    maxDepth = get_max_depth(nestedList, 1)
    
    # Second pass to compute the inverse weighted sum
    def helper(lst, depth):
        total = 0
        for ni in lst:
            if ni.isInteger():
                # weight = maxDepth - depth + 1
                total += ni.getInteger() * (maxDepth - depth + 1)
            else:
                total += helper(ni.getList(), depth+1)
        return total
    return helper(nestedList, 1)

This two-pass method is clear: the first traversal finds how deep the nesting goes, and the second does the summation with appropriate weights. Its complexity is O(n) time and O(d) space, similar to the original problem (technically O(2n) ~ O(n) time due to two passes). Given the constraints, two passes are not a problem.

Optimized Approach (Single-Pass): We can actually solve the inverse weight sum in one pass using a clever BFS strategy. The trick is to accumulate the sum of integers at each depth in a running manner. Specifically:

• Traverse the nested list level by level (BFS).
• Keep track of two sums: an unweighted sum (sum of integers seen so far without weight) and a weighted sum (which will accumulate the final result).
• At each level of BFS, add all integers at that level to unweighted. After processing that level, add the entire unweighted sum to weighted.
• Move to the next level and repeat.

Why does this work? In the end, unweighted will have the sum of all integers (because we keep adding all integers level by level), and weighted effectively adds the sum of integers once for each level from the bottom. An integer that is at depth d will be added to weighted in each level of the BFS from level d up to maxDepth. That means it gets added (maxDepth - d + 1) times, which is exactly its weight in the inverted scheme.

For example, using this method on [1,[4,[6]]]:

• Start with unweighted = 0, weighted = 0.
• Level 1: elements = [1, sublist]. unweighted += 1. After level, weighted += unweighted (now weighted = 1). Enqueue sublist.
• Level 2: elements = [4, sub-sublist]. unweighted += 4 (unweighted was 1, now becomes 5). After level, weighted += unweighted (now weighted = 1 + 5 = 6). Enqueue sub-sublist.
• Level 3: elements = [6]. unweighted += 6 (was 5, now 11 which is sum of all numbers). After level, weighted += unweighted (weighted = 6 + 11 = 17). Done. weighted is 17 which matches the answer.

Here is a Python implementation of the one-pass BFS approach:

from collections import deque

def depthSumInverse_one_pass(nestedList):
    queue = deque(nestedList)
    unweighted = 0
    weighted = 0
    while queue:
        # process one level at a time
        for _ in range(len(queue)):
            ni = queue.popleft()
            if ni.isInteger():
                unweighted += ni.getInteger()
            else:
                for child in ni.getList():
                    queue.append(child)
        # after processing this level, add the running sum (unweighted) to weighted result
        weighted += unweighted
    return weighted

In this code, unweighted accumulates the sum of all integers seen so far (by the time we've gone through the whole list, unweighted is just the sum of all integers). weighted is increased by the current unweighted after each level, effectively giving each integer an extra count for each level above it. The result is the inverse-weighted sum. The time complexity is O(n) and space O(n), similar to other approaches.

Both the two-pass and single-pass methods will yield the correct result for the inverse weight problem. The single-pass BFS is a neat trick specific to the inverse weighting scenario, leveraging the structure of BFS traversal to implicitly calculate weights.

Related Problems for Further Practice

To strengthen understanding of nested structures, depth calculations, and recursion, you can practice the following related problems:

• Flatten Nested List Iterator (LeetCode 341) – Implement an iterator that flattens a nested list of integers. This problem uses the same NestedInteger interface and requires managing a stack/recursion to flatten on the fly.
• Nested List Weight Sum II (LeetCode 364) – The variation discussed above with reverse depth weights. It’s a direct follow-up to try out the two-pass and one-pass solutions described.
• Mini Parser (LeetCode 385) – Parse a string of a nested list and produce the NestedInteger object. This involves recursion/stack to build a nested structure from a string, which is good practice for understanding nested data representations.
• Employee Importance (LeetCode 690) – You’re given a data structure of employees with IDs, importance values, and subordinates (which form an organizational hierarchy). The task is to sum up an employee’s importance value and those of all their direct and indirect subordinates. This can be solved with DFS or BFS on a nested-like structure (graph traversal).
• Maximum Depth of Binary Tree (LeetCode 104) – Although this is about a binary tree, it’s a simpler recursion problem for depth. It helps practice the idea of depth calculation in a hierarchical structure (here, a tree) and can be a stepping stone to dealing with nested lists or more complex recursive structures.