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Nested List Weight Sum

Medium

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Topics:

Recursion

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.

The depth of an integer is the number of lists that it is inside of. For example, the nested list [1,[2,2],[[3],2],1] has depth values of 1, 2, 3 for integers 1, 2, 3 respectively.

Return the sum of each integer in nestedList multiplied by its depth.

Example 1:

Input: nestedList = [[1,1],2,[1,1]]
Output: 10
Explanation: Because there are four 1's at depth 2, and one 2 at depth 1.
1*2 + 1*2 + 2*1 + 1*2 + 1*2 = 10

Example 2:

Input: nestedList = [1,[4,[6]]]
Output: 27
Explanation: One 1 at depth 1, one 4 at depth 2, and one 6 at depth 3.
1*1 + 4*2 + 6*3 = 27

Example 3:

Input: nestedList = [0]
Output: 0

Constraints:

1 <= nestedList.length <= 50
The depth of the nested lists is at most 50.

Nested List Weight Sum

Medium

0 views

Topics:

Recursion

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.

The depth of an integer is the number of lists that it is inside of. For example, the nested list [1,[2,2],[[3],2],1] has depth values of 1, 2, 3 for integers 1, 2, 3 respectively.

Return the sum of each integer in nestedList multiplied by its depth.

Example 1:

Input: nestedList = [[1,1],2,[1,1]]
Output: 10
Explanation: Because there are four 1's at depth 2, and one 2 at depth 1.
1*2 + 1*2 + 2*1 + 1*2 + 1*2 = 10

Example 2:

Input: nestedList = [1,[4,[6]]]
Output: 27
Explanation: One 1 at depth 1, one 4 at depth 2, and one 6 at depth 3.
1*1 + 4*2 + 6*3 = 27

Example 3:

Input: nestedList = [0]
Output: 0

Constraints:

1 <= nestedList.length <= 50
The depth of the nested lists is at most 50.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the nested lists contain null or empty lists?
What is the maximum depth of the nested list?
Can the integers within the nested lists be negative, zero, or only positive?
If the input `nestedList` is empty or null, what should I return?
Are the elements within a list guaranteed to be either an integer or another list, or could there be other data types?

Brute Force Solution

Approach

The brute force approach calculates the weighted sum of a nested list by exploring every possible depth of each element. We essentially go through each number or sub-list and keep track of its 'level' within the overall structure. Then we multiply each number by its level and add everything up.

Here's how the algorithm would work step-by-step:

Start at the beginning of the list.
If you find a number, multiply it by its current level (which starts at 1) and add it to a running total.
If you find another list inside the main list, increase the level by one.
Go through each item in the sub-list and repeat the above process.
Once you've gone through every item in the sub-list, go back to the previous level.
Continue going through the main list (or any other sub-list) until you've checked every item.
The final running total is the weighted sum.

Code Implementation

def nested_list_weight_sum(nested_list):
    total_sum = 0
    
    def calculate_weight_sum(nested_list, depth):
        nonlocal total_sum

        for element in nested_list:
            if isinstance(element, int):
                # Numbers are multiplied by depth
                total_sum += element * depth

            elif isinstance(element, list):
                # Recurse into sublists with increased depth
                calculate_weight_sum(element, depth + 1)

    # Start the calculation with initial depth of 1
    calculate_weight_sum(nested_list, 1)

    return total_sum

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(N) – Let N be the total number of integers within the nested list structure. The algorithm visits each integer exactly once to calculate its weighted sum. When a sublist is encountered, the algorithm explores its elements, ensuring each integer is processed regardless of its depth. Therefore, the time complexity is directly proportional to the total number of integers, N, across all levels of nesting. This results in a linear time complexity of O(N).

Space Complexity

O(D) – The algorithm uses recursion to explore the nested lists. The space complexity is determined by the maximum depth of nesting, denoted as D, because each recursive call adds a new frame to the call stack. In the worst case, the depth could be significant, leading to D stack frames. Therefore, the auxiliary space used is proportional to the maximum depth of the nested lists.

Optimal Solution

Approach

The goal is to calculate the sum of numbers within a special list structure, where each number's value is multiplied by its depth within the list. We can use a technique to explore the list structure, keeping track of the current depth, and adding the correctly weighted values as we find the numbers.

Here's how the algorithm would work step-by-step:

Imagine you're exploring a tree. You start at the root (the outermost list).
Keep track of how deep you are in the tree (the current depth). Start at depth 1.
As you go through the list, if you find a number, multiply that number by your current depth and add it to a running total.
If you find another list inside the list, increase your current depth by one and explore that inner list using the same method.
Once you've gone through the inner list, decrease your current depth by one when you go back up to the outer list.
Continue this process until you've explored all the lists and numbers.
The final total will be the weighted sum of all the numbers.

Code Implementation

def nested_list_weight_sum(nested_list):
    return calculate_weight_sum(nested_list, 1)

def calculate_weight_sum(nested_list, depth):
    total_sum = 0

    for element in nested_list:
        if isinstance(element, int):
            # Element is an integer, so add to the weighted sum
            total_sum += element * depth

        elif isinstance(element, list):
            # Element is a list, so recursively calculate the sum
            total_sum += calculate_weight_sum(element, depth + 1)

    return total_sum

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The algorithm traverses the nested list structure using depth-first search. Each integer within the nested list is visited and processed exactly once. Therefore, the time complexity is directly proportional to the total number of integers and lists in the input, which we can represent as n. Thus, the overall time complexity is O(n).

Space Complexity

O(D) – The primary driver of auxiliary space is the recursion depth. Each time the algorithm encounters a nested list, it makes a recursive call, increasing the depth by one. In the worst-case scenario, the lists are nested to a maximum depth D, resulting in D stack frames being added to the call stack. Therefore, the auxiliary space used is proportional to the maximum nesting depth of the input list, which is O(D).

Edge Cases

Case	How to Handle
Null or empty nestedList input	Return 0 immediately as there are no integers to sum.
Nested list contains only empty lists	Return 0 as the empty lists contain no integers.
Deeply nested list that could exceed recursion depth limits.	Iterative solution using a stack prevents stack overflow errors from excessive recursion.
Nested list contains extremely large integers.	Use a 64-bit integer type (long) to prevent integer overflow when calculating the weighted sum.
Nested list contains both extremely large and extremely small integers (close to integer limits).	Ensure calculations with different sized integers does not overflow or underflow.
A single integer as the entire input list.	Correctly handle the single integer with depth 1.
Nested lists of varying depths, some with very large depth.	The depth variable will track the current depth of recursion, ensuring it's accurately multiplied with integer values.
Nested lists containing zero values.	Zero values are correctly handled as integers with a corresponding depth.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the nested lists contain null or empty lists?
What is the maximum depth of the nested list?
Can the integers within the nested lists be negative, zero, or only positive?
If the input `nestedList` is empty or null, what should I return?
Are the elements within a list guaranteed to be either an integer or another list, or could there be other data types?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start at the beginning of the list.
If you find a number, multiply it by its current level (which starts at 1) and add it to a running total.
If you find another list inside the main list, increase the level by one.
Go through each item in the sub-list and repeat the above process.
Once you've gone through every item in the sub-list, go back to the previous level.
Continue going through the main list (or any other sub-list) until you've checked every item.
The final running total is the weighted sum.

Code Implementation

def nested_list_weight_sum(nested_list):
    total_sum = 0
    
    def calculate_weight_sum(nested_list, depth):
        nonlocal total_sum

        for element in nested_list:
            if isinstance(element, int):
                # Numbers are multiplied by depth
                total_sum += element * depth

            elif isinstance(element, list):
                # Recurse into sublists with increased depth
                calculate_weight_sum(element, depth + 1)

    # Start the calculation with initial depth of 1
    calculate_weight_sum(nested_list, 1)

    return total_sum

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Imagine you're exploring a tree. You start at the root (the outermost list).
Keep track of how deep you are in the tree (the current depth). Start at depth 1.
As you go through the list, if you find a number, multiply that number by your current depth and add it to a running total.
If you find another list inside the list, increase your current depth by one and explore that inner list using the same method.
Once you've gone through the inner list, decrease your current depth by one when you go back up to the outer list.
Continue this process until you've explored all the lists and numbers.
The final total will be the weighted sum of all the numbers.

Code Implementation

def nested_list_weight_sum(nested_list):
    return calculate_weight_sum(nested_list, 1)

def calculate_weight_sum(nested_list, depth):
    total_sum = 0

    for element in nested_list:
        if isinstance(element, int):
            # Element is an integer, so add to the weighted sum
            total_sum += element * depth

        elif isinstance(element, list):
            # Element is a list, so recursively calculate the sum
            total_sum += calculate_weight_sum(element, depth + 1)

    return total_sum

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty nestedList input	Return 0 immediately as there are no integers to sum.
Nested list contains only empty lists	Return 0 as the empty lists contain no integers.
Deeply nested list that could exceed recursion depth limits.	Iterative solution using a stack prevents stack overflow errors from excessive recursion.
Nested list contains extremely large integers.	Use a 64-bit integer type (long) to prevent integer overflow when calculating the weighted sum.
Nested list contains both extremely large and extremely small integers (close to integer limits).	Ensure calculations with different sized integers does not overflow or underflow.
A single integer as the entire input list.	Correctly handle the single integer with depth 1.
Nested lists of varying depths, some with very large depth.	The depth variable will track the current depth of recursion, ensuring it's accurately multiplied with integer values.
Nested lists containing zero values.	Zero values are correctly handled as integers with a corresponding depth.