Problem Description

You are given a nested list of integers where each element can be either:

• An integer
• A list containing integers or other lists (which can be nested further)

The depth of an integer is determined by how many lists it is nested inside. For example, in the nested list [1,[2,2],[[3],2],1]:

• The first 1 and last 1 are at depth 1 (inside one list)
• The two 2s in [2,2] are at depth 2 (inside two lists)
• The 3 in [[3],2] is at depth 3 (inside three lists)
• The 2 in [[3],2] is at depth 2 (inside two lists)

The weight of an integer is calculated using the formula: weight = maxDepth - depth + 1, where maxDepth is the maximum depth of any integer in the entire nested list.

Your task is to calculate the sum of all integers in the nested list, where each integer is multiplied by its weight.

For example, if maxDepth = 3 for the list [1,[2,2],[[3],2],1]:

• The integers at depth 1 have weight = 3 - 1 + 1 = 3
• The integers at depth 2 have weight = 3 - 2 + 1 = 2
• The integers at depth 3 have weight = 3 - 3 + 1 = 1

The final result would be: 1×3 + 2×2 + 2×2 + 3×1 + 2×2 + 1×3 = 3 + 4 + 4 + 3 + 4 + 3 = 21

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The nested list structure can be viewed as a graph/tree where each list is a node that can contain integers (leaf nodes) or other lists (child nodes). This hierarchical structure represents a tree-like graph.

Is it a tree?

• Yes: The nested list structure is indeed a tree. Each nested list represents a node in the tree, with integers as leaf nodes and sublists as internal nodes. There's a clear parent-child relationship without cycles, making it a tree structure.

DFS

• Yes: We arrive at DFS as our algorithm choice.

Conclusion: The flowchart suggests using a Depth-First Search (DFS) approach for this problem.

This makes perfect sense because:

1. We need to traverse through all levels of nesting to find the maximum depth
2. We need to visit every integer in the nested structure while tracking its depth
3. DFS naturally handles the recursive nature of nested lists, allowing us to dive deep into each nested level before backtracking
4. The depth tracking during DFS traversal directly corresponds to counting how many lists each integer is nested within

The DFS approach allows us to:

• Calculate the depth of each integer as we traverse
• Keep track of the maximum depth encountered
• Accumulate the sum of integers and their weighted sums in a single pass
• Handle the recursive structure of nested lists naturally

Intuition

The key insight is recognizing that calculating weights directly would require two passes through the nested list: first to find the maximum depth, then to calculate the weighted sum. But we can be clever about this.

Let's think about what we're actually computing. For each integer with value v at depth d, its contribution to the final sum is v × (maxDepth - d + 1). If we expand this:

• v × (maxDepth - d + 1)
• v × maxDepth - v × d + v

If we sum this across all integers, we get:

• (sum of all v × maxDepth) - (sum of all v × d) + (sum of all v)
• maxDepth × (sum of all v) - (sum of all v × d) + (sum of all v)
• (maxDepth + 1) × (sum of all v) - (sum of all v × d)

This transformation is brilliant because it separates our calculation into two parts:

1. s: The simple sum of all integers
2. ws: The weighted sum where each integer is multiplied by its depth

During a single DFS traversal, we can:

• Track the current depth as we descend into nested lists
• Update the maximum depth whenever we encounter an integer
• Add each integer to our simple sum s
• Add each integer multiplied by its current depth to our weighted sum ws

After the traversal, we have everything we need: maxDepth, s, and ws. The final answer is simply (maxDepth + 1) × s - ws.

This approach transforms what seems like a two-pass problem (find max depth, then calculate weighted sum) into an elegant single-pass solution using mathematical manipulation.

Solution Approach

The implementation uses DFS with three tracking variables: maxDepth, s (simple sum), and ws (weighted sum by depth).

DFS Function Design:

We create a helper function dfs(x, d) that processes a NestedInteger object x at depth d:

1. Update Maximum Depth: First, we update maxDepth = max(maxDepth, d) to track the deepest level we've encountered.

2. Handle Base Case (Integer): If x.isInteger() returns True:

   • Add the integer value to our simple sum: s += x.getInteger()
   • Add the depth-weighted value to our weighted sum: ws += x.getInteger() * d

3. Handle Recursive Case (List): If x is a list:

   • Iterate through each element y in x.getList()
   • Recursively call dfs(y, d + 1) to process each element at the next depth level

Main Algorithm:

1. Initialize our tracking variables: maxDepth = s = ws = 0

2. Process the input nestedList:

   • For each top-level element x in nestedList
   • Call dfs(x, 1) starting at depth 1

3. Calculate the final result using our mathematical formula:

   • Return (maxDepth + 1) * s - ws

Why This Works:

The algorithm leverages the mathematical transformation we derived:

• Instead of calculating weight = maxDepth - depth + 1 for each integer after finding maxDepth
• We collect s (sum of all integers) and ws (sum of integers × their depths) in one pass
• Then apply the formula (maxDepth + 1) × s - ws to get the same result

Time and Space Complexity:

• Time: O(n) where n is the total number of elements (integers and lists) in the nested structure
• Space: O(d) where d is the maximum depth, for the recursion call stack

Example Walkthrough

Let's walk through the solution with the nested list [[1,1],2,[1,1]].

Step 1: Visualize the Structure

[[1,1],2,[1,1]]
├── [1,1]      (depth 1)
│   ├── 1      (depth 2)
│   └── 1      (depth 2)
├── 2          (depth 1)
└── [1,1]      (depth 1)
    ├── 1      (depth 2)
    └── 1      (depth 2)

Step 2: DFS Traversal

Initialize: maxDepth = 0, s = 0, ws = 0

Process first element [1,1] at depth 1:

• It's a list, so recurse into its elements
• Process first 1 at depth 2:
  • Update: maxDepth = 2
  • Add to simple sum: s = 0 + 1 = 1
  • Add to weighted sum: ws = 0 + (1 × 2) = 2
• Process second 1 at depth 2:
  • maxDepth remains 2
  • Update: s = 1 + 1 = 2
  • Update: ws = 2 + (1 × 2) = 4

Process second element 2 at depth 1:

• It's an integer
• maxDepth remains 2
• Update: s = 2 + 2 = 4
• Update: ws = 4 + (2 × 1) = 6

Process third element [1,1] at depth 1:

• It's a list, so recurse into its elements
• Process first 1 at depth 2:
  • maxDepth remains 2
  • Update: s = 4 + 1 = 5
  • Update: ws = 6 + (1 × 2) = 8
• Process second 1 at depth 2:
  • maxDepth remains 2
  • Update: s = 5 + 1 = 6
  • Update: ws = 8 + (1 × 2) = 10

Step 3: Apply the Formula

After traversal, we have:

• maxDepth = 2
• s = 6 (sum of all integers: 1+1+2+1+1)
• ws = 10 (weighted sum by depth: 1×2 + 1×2 + 2×1 + 1×2 + 1×2)

Final result: (maxDepth + 1) × s - ws = (2 + 1) × 6 - 10 = 18 - 10 = 8

Step 4: Verify with Direct Calculation

Let's verify by calculating weights directly:

• Integers at depth 1 have weight = 2 - 1 + 1 = 2
• Integers at depth 2 have weight = 2 - 2 + 1 = 1

Weighted sum:

• Four integers (1,1,1,1) at depth 2: 4 × 1 = 4
• One integer (2) at depth 1: 2 × 2 = 4
• Total: 4 + 4 = 8 ✓

The formula correctly transforms the two-pass weight calculation into a single-pass algorithm by collecting the simple sum and depth-weighted sum during traversal.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm performs a depth-first search (DFS) traversal through all elements in the nested list structure. Each integer in the nested structure is visited exactly once during the traversal. For each integer encountered, the algorithm performs constant-time operations:

• Updating maxDepth with a max comparison
• Adding the integer value to the sum s
• Adding the weighted value to ws

Since we visit each of the n integers exactly once and perform O(1) operations for each, the overall time complexity is O(n), where n is the total number of integers in the nested structure.

Space Complexity: O(d)

The space complexity is determined by the recursive call stack of the DFS function. The maximum depth of recursion equals the maximum nesting depth d of the input structure. Each recursive call adds a new frame to the call stack, consuming O(1) space for the function parameters and local variables. Therefore, the worst-case space complexity is O(d), where d is the maximum depth of nesting.

Note: The reference answer states O(n) for space complexity, which would be accurate if considering the worst-case scenario where the nested structure forms a linear chain (like [[[[[5]]]]]), making the depth equal to n. In general practice, the space complexity is more precisely described as O(d) where d is the maximum nesting depth.

Common Pitfalls

1. Incorrect Initial Depth Value

One of the most common mistakes is starting the traversal with depth 0 instead of depth 1. The problem states that elements in the outermost list are at depth 1, not depth 0.

Incorrect Implementation:

# Wrong: Starting with depth 0
for nested_element in nestedList:
    traverse_nested_list(nested_element, 0)  # ❌ Should be 1

Correct Implementation:

# Correct: Starting with depth 1
for nested_element in nestedList:
    traverse_nested_list(nested_element, 1)  # ✓

2. Misunderstanding the NestedInteger Interface

Developers often confuse how to handle the NestedInteger interface, particularly when dealing with empty lists or assuming the wrong return types.

Common Mistakes:

# Wrong: Assuming getList() returns integers directly
for element in nested_integer.getList():
    total_sum += element  # ❌ element is NestedInteger, not int

# Wrong: Not checking if it's an integer before calling getInteger()
value = nested_integer.getInteger()  # ❌ May fail if it's a list

Correct Approach:

if nested_integer.isInteger():
    value = nested_integer.getInteger()
    # Process integer value
else:
    for nested_element in nested_integer.getList():
        # Process each NestedInteger recursively

3. Formula Application Error

The mathematical formula (maxDepth + 1) * s - ws might be incorrectly implemented, especially with off-by-one errors.

Incorrect Implementations:

# Wrong formulas that developers might use:
return maxDepth * total_sum - weighted_sum_by_depth  # ❌ Missing +1
return (maxDepth - 1) * total_sum + weighted_sum_by_depth  # ❌ Wrong signs
return total_sum * maxDepth - weighted_sum_by_depth + total_sum  # ❌ Expanded incorrectly

Correct Formula:

return (max_depth + 1) * total_sum - weighted_sum_by_depth  # ✓

4. Variable Scope Issues with Nested Functions

Forgetting to use nonlocal for variables that need to be modified within the nested function.

Incorrect Implementation:

def depthSumInverse(self, nestedList):
    max_depth = 0
    total_sum = 0
    weighted_sum_by_depth = 0
  
    def traverse_nested_list(nested_integer, current_depth):
        # ❌ Without nonlocal, these create new local variables
        max_depth = max(max_depth, current_depth)  # UnboundLocalError
        total_sum += value  # UnboundLocalError

Correct Implementation:

def traverse_nested_list(nested_integer, current_depth):
    nonlocal max_depth, total_sum, weighted_sum_by_depth  # ✓
    max_depth = max(max_depth, current_depth)
    # ... rest of the function

5. Handling Edge Cases

Not properly handling edge cases like empty lists or deeply nested empty structures.

Potential Issues:

• Empty input list [] should return 0
• Nested empty lists like [[[]]] should not affect the result
• Single integer wrapped in multiple lists [[[5]]] should be handled correctly

Robust Solution:

# Handle empty input
if not nestedList:
    return 0

# The existing algorithm naturally handles empty nested lists
# since they don't contain integers to add to the sums