Nested List Weight Sum II

Medium

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Topics:

ArraysRecursion

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 each integer's value set to its depth.

Return the sum of each integer in nestedList multiplied by its depth. Note: You should not multiply by the depth.

Return the sum of each integer in nestedList multiplied by its maximum depth level. That is, if the maximum depth of the nested list is depth, then decrement each integer's depth by one before multiplying the integer by depth.

Example 1:

Input: nestedList = [[1,1],2,[1,1]]
Output: 8
Explanation: Four 1's at depth 1, one 2 at depth 2.
1*2 + 1*2 + 2*1 + 1*2 + 1*2 = 8

Example 2:

Input: nestedList = [1,[4,[6]]]
Output: 17
Explanation: One 1 at depth 3, one 4 at depth 2, and one 6 at depth 1.
1*3 + 4*2 + 6*1 = 17

Constraints:

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

Try coding it in LeetCode

Nested List Weight Sum II

Medium

1 view

Topics:

ArraysRecursion

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 each integer's value set to its depth.

Return the sum of each integer in nestedList multiplied by its depth. Note: You should not multiply by the depth.

Return the sum of each integer in nestedList multiplied by its maximum depth level. That is, if the maximum depth of the nested list is depth, then decrement each integer's depth by one before multiplying the integer by depth.

Example 1:

Input: nestedList = [[1,1],2,[1,1]]
Output: 8
Explanation: Four 1's at depth 1, one 2 at depth 2.
1*2 + 1*2 + 2*1 + 1*2 + 1*2 = 8

Example 2:

Input: nestedList = [1,[4,[6]]]
Output: 17
Explanation: One 1 at depth 3, one 4 at depth 2, and one 6 at depth 1.
1*3 + 4*2 + 6*1 = 17

Constraints:

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

Try coding it in LeetCode

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 list contain null or empty lists at any level?
What is the range of integer values within the nested lists? Should I be concerned about integer overflow when calculating the weighted sum?
Could you provide an example of a deeply nested list to illustrate how the bottom-up depth is calculated?
Is the input `nestedList` guaranteed to be a valid nested list, or do I need to handle cases where it might be malformed (e.g., containing elements that are neither integers nor lists)?
If the `nestedList` is empty, should I return 0 or is there a specific error condition I need to handle?

Brute Force Solution

Approach

The brute force approach involves figuring out the depth of each number within the nested structure. We'll determine the maximum depth first, then use that to weight each number based on its level of nesting.

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

First, find the maximum depth of the nested structure. This means going as deep as possible within the lists to find the level of the most deeply nested number.
Once we know the maximum depth, we go through the nested structure again.
For each number we find, we calculate its weight by subtracting its depth from the maximum depth and adding one. Deeper numbers get smaller weights.
Multiply each number by its calculated weight.
Finally, add up all the weighted numbers to get the total weighted sum.

Code Implementation

def nested_list_weight_sum_two(nested_list):
    def get_depth(nested_list):
        maximum_depth_so_far = 1
        for element in nested_list:
            if isinstance(element, list):
                maximum_depth_so_far = max(maximum_depth_so_far, 1 + get_depth(element))
        return maximum_depth_so_far

    maximum_depth = get_depth(nested_list)

    # Need maximum depth to apply weights

    def calculate_weighted_sum(nested_list, current_depth):
        weighted_sum = 0
        for element in nested_list:
            if isinstance(element, list):

                weighted_sum += calculate_weighted_sum(element, current_depth + 1)

            else:
                weight = maximum_depth - current_depth + 1

                weighted_sum += element * weight

        return weighted_sum

    # Start the recursive calculation

    return calculate_weighted_sum(nested_list, 1)

Big(O) Analysis

Time Complexity

O(n) – The algorithm traverses the nested list structure twice. The first traversal calculates the maximum depth, which requires visiting each element in the nested list once. The second traversal calculates the weighted sum, also requiring a visit to each element. Therefore, the time complexity is proportional to the number of elements in the nested list, denoted as n, making it O(n).

Space Complexity

O(D) – The space complexity is determined primarily by the maximum recursion depth (D) during the depth calculation phase. The recursive calls to determine the maximum depth create stack frames on the call stack. The maximum number of stack frames, and hence the auxiliary space used, corresponds to the maximum depth of nested lists, where D is the maximum depth. Therefore, the space complexity is O(D).

Optimal Solution

Approach

This problem asks us to calculate a weighted sum of integers within a nested list structure, where the weight is determined by the depth of the integer from the outermost list. The key idea is to traverse the nested list level by level, accumulating the unweighted sum of the integers at each level and then multiplying the cumulative sum by the level's depth.

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

First, we need to figure out how deep the nested lists go. Find the maximum depth of the structure.
Instead of calculating the weighted sum directly, let's go level by level.
At each level, find all the integers and add them together to get the sum for that level.
Keep track of these level sums as you go. Store them in a temporary storage.
Once you've gone through all levels, you have a collection of sums representing each level's total.
Now, you can determine the overall sum using the total depth of the structure and the level sums by multiplying the level sum by how deep it is in the nested structure.
Add up all of these weighted level sums to get the final result, which is the weighted sum of all integers in the nested list.

Code Implementation

def nested_list_weight_sum(nested_list):

    def get_depth(nested_list_item):
        if not isinstance(nested_list_item, list):
            return 1
        if not nested_list_item:
            return 1
        return 1 + max(get_depth(item) for item in nested_list_item)

    maximum_depth = get_depth(nested_list)

    level_sums = []

    def calculate_level_sums(nested_list_item, depth):
        if len(level_sums) < depth:
            level_sums.append(0)

        for item in nested_list_item:
            if isinstance(item, int):
                level_sums[depth - 1] += item
            elif isinstance(item, list):
                calculate_level_sums(item, depth + 1)

    # Calculating sum for each level
    calculate_level_sums(nested_list, 1)

    total_weighted_sum = 0

    # Multiplying each level sum by depth
    for index, level_sum in enumerate(level_sums):
        total_weighted_sum += level_sum * (maximum_depth - index)

    return total_weighted_sum

Big(O) Analysis

Time Complexity

O(n) – Let n be the total number of elements (integers and lists) within the nested list. The algorithm involves traversing the nested list structure to find the maximum depth, which takes O(n) time. Then, it iterates through the list level by level, summing the integers at each level, also taking O(n) time in total across all levels since each element is visited. Finally, it calculates the weighted sum based on the level sums, which takes O(d) time, where d is the maximum depth. Since d <= n, the dominant factor is O(n).

Space Complexity

O(N) – The algorithm uses temporary storage to keep track of the level sums. In the worst-case scenario, where each level contains a significant number of integers or a single integer, the space required to store these level sums could grow linearly with the maximum depth of the nested lists, represented by N (where N is the number of elements across all lists). Therefore, the auxiliary space is proportional to N. The level-by-level traversal creates an unweighted level sum, and this sum of sums is weighted by the depth. So we store intermediate results in sums.

Edge Cases

Case	How to Handle
Null or empty nestedList input	Return 0 immediately as there are no integers to sum.
Nested list containing only empty lists	Return 0 because the weighted sum will be zero due to no integers.
Nested lists with extremely deep nesting levels	Ensure the recursion depth or iterative depth calculation doesn't lead to stack overflow or performance issues; consider an iterative approach if recursion depth becomes a concern.
Nested list containing very large integer values (potential overflow)	Use appropriate data types (e.g., long) to avoid integer overflow when calculating the weighted sum.
Nested list with integers having zero value	The zero values should contribute zero to the sum, and should be handled correctly by the weighting calculation.
Nested List with only single integer at root level	Correctly compute depth and the single integer will be multiplied by the max depth.
Nested Lists with varying depth distributions	The depth calculation and the weighted sum must correctly handle skewed depth distributions, ensuring that leaves have depth 1 and the root has the largest depth.
Maximum sized nested list (memory considerations)	Iterative implementations will use less memory than recursive ones for large input lists.