Find Products of Elements of Big Array

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ArraysBit Manipulation

The powerful array of a non-negative integer x is defined as the shortest sorted array of powers of two that sum up to x. The table below illustrates examples of how the powerful array is determined. It can be proven that the powerful array of x is unique.

num	Binary Representation	powerful array
1	00001	[1]
8	01000	[8]
10	01010	[2, 8]
13	01101	[1, 4, 8]
23	10111	[1, 2, 4, 16]

The array big_nums is created by concatenating the powerful arrays for every positive integer i in ascending order: 1, 2, 3, and so on. Thus, big_nums begins as [1, 2, 1, 2, 4, 1, 4, 2, 4, 1, 2, 4, 8, ...].

You are given a 2D integer matrix queries, where for queries[i] = [from_i, to_i, mod_i] you should calculate (big_nums[from_i] * big_nums[from_i + 1] * ... * big_nums[to_i]) % mod_i.

Return an integer array answer such that answer[i] is the answer to the i^th query.

Example 1:

Input: queries = [[1,3,7]]

Output: [4]

Explanation:

There is one query.

big_nums[1..3] = [2,1,2]. The product of them is 4. The result is 4 % 7 = 4.

Example 2:

Input: queries = [[2,5,3],[7,7,4]]

Output: [2,2]

Explanation:

There are two queries.

First query: big_nums[2..5] = [1,2,4,1]. The product of them is 8. The result is 8 % 3 = 2.

Second query: big_nums[7] = 2. The result is 2 % 4 = 2.

Constraints:

1 <= queries.length <= 500
queries[i].length == 3
0 <= queries[i][0] <= queries[i][1] <= 10¹⁵
1 <= queries[i][2] <= 10⁵

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:

What is the maximum possible size of the input array `nums`?
Can the input array `nums` contain negative numbers, zero, or floating-point numbers?
What should I return if the product is too large to be represented as a standard integer (overflow)?
Is an empty input array a possibility, and if so, what value should I return?
Are there any specific error conditions (e.g., null input) that I need to handle?

Brute Force Solution

Approach

We want to find the product of all numbers, except for the number at each position. The brute force method involves calculating the product for each position separately, without any clever shortcuts.

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

For the first number, multiply all the other numbers together.
For the second number, multiply all the numbers together, skipping the second number.
Do this for every number in the list: multiply all the other numbers together, leaving out the number at the current position.
The result from each of these calculations is the answer for that position.

Code Implementation

def products_of_elements_big_array(numbers):
    results = []
    for index in range(len(numbers)):
        product = 1
        # Calculate product excluding the current number

        for second_index in range(len(numbers)):
            if index != second_index:
                product *= numbers[second_index]

        # Store product of all other numbers

        results.append(product)
    return results

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n elements in the input array. For each element, it calculates the product of all other elements, requiring a traversal of the remaining (n-1) elements. This results in approximately n * (n-1) multiplications. Therefore, the total number of operations grows proportionally to n squared, making the time complexity O(n²).

Space Complexity

O(1) – The provided plain English explanation describes calculating the product for each position separately *without* storing intermediate products across iterations. This implies no auxiliary data structures are needed to store intermediate results, only potentially a variable to hold the running product for a single element. Therefore, the algorithm uses a constant amount of extra space regardless of the input array size N, resulting in O(1) space complexity.

Optimal Solution

Approach

The most efficient way to find the products of all elements in a big list, excluding each element one at a time, is to avoid redundant multiplication. We achieve this by calculating prefixes and suffixes of the list and then multiplying them together to obtain the result for each element efficiently.

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

First, imagine calculating the product of all numbers from the beginning of the list up to each position. Store these results.
Next, do something similar but start from the end of the list and calculate the products going backward. Store these backward products too.
Now, for each position in the list, the product we want (excluding the number at that position) can be found by multiplying the prefix product just before that position by the suffix product just after that position.
For the very first number in the list, there is no prefix, so we just use the suffix starting from the second number.
Similarly, for the last number, we use only the prefix ending at the second-to-last number.
By using these prefix and suffix products, we efficiently calculate the products without having to redo the multiplication for each number separately.

Code Implementation

def find_products_of_elements(number_list):
    list_length = len(number_list)
    prefix_products = [1] * list_length
    suffix_products = [1] * list_length

    # Calculate prefix products.
    for i in range(1, list_length):
        prefix_products[i] = prefix_products[i - 1] * number_list[i - 1]

    # Calculate suffix products.
    for i in range(list_length - 2, -1, -1):
        suffix_products[i] = suffix_products[i + 1] * number_list[i + 1]

    result = [1] * list_length

    # Compute the final products using prefix and suffix products.
    for i in range(list_length):
        if i == 0:
            result[i] = suffix_products[i]

        elif i == list_length - 1:
            result[i] = prefix_products[i]

        else:
            result[i] = prefix_products[i] * suffix_products[i]

    return result

Big(O) Analysis

Time Complexity

O(n) – The algorithm calculates prefix products by iterating through the array once, taking O(n) time. Similarly, it calculates suffix products in another pass, also taking O(n) time. Finally, it iterates through the array a third time to compute the final products using the prefix and suffix arrays, which again takes O(n) time. Since these are sequential operations, the total time complexity is O(n) + O(n) + O(n), which simplifies to O(n).

Space Complexity

O(N) – The algorithm creates two auxiliary arrays, one for storing prefix products and another for storing suffix products. Both arrays have the same length as the input list, denoted as N. Therefore, the extra space used is directly proportional to the size of the input, requiring space for 2N elements. This simplifies to O(N) space complexity.

Edge Cases

Case	How to Handle
Null or empty input array	Return 1, as the product of no elements is defined as 1, or throw an IllegalArgumentException if null input is not permitted.
Array contains a zero	Return 0 immediately, as any product involving zero is zero.
Array contains very large positive numbers that might cause integer overflow	Use long data type for intermediate product calculations, or utilize libraries that handle arbitrarily large numbers.
Array contains very small negative numbers that might cause integer overflow	Use long data type for intermediate product calculations, or utilize libraries that handle arbitrarily large numbers.
Array contains both very large positive and small negative numbers	Use long data type for intermediate product calculations, or utilize libraries that handle arbitrarily large numbers.
Array with only one element	Return the element itself, as it is the product of all elements in the array.
Array with all identical values	The product will be value^n, which might trigger overflow issues, handled with long or arbitrary precision libraries.
An extremely large array to assess scalability	Iterative approach with long datatype should work, but consider alternatives if more memory optimization is needed.

Null or empty input array

How to Handle:

Return 1, as the product of no elements is defined as 1, or throw an IllegalArgumentException if null input is not permitted.

Array contains a zero

How to Handle:

Return 0 immediately, as any product involving zero is zero.

Array contains very large positive numbers that might cause integer overflow

How to Handle:

Use long data type for intermediate product calculations, or utilize libraries that handle arbitrarily large numbers.

Array contains very small negative numbers that might cause integer overflow

How to Handle:

Use long data type for intermediate product calculations, or utilize libraries that handle arbitrarily large numbers.

Array contains both very large positive and small negative numbers

How to Handle:

Use long data type for intermediate product calculations, or utilize libraries that handle arbitrarily large numbers.

Array with only one element

How to Handle:

Return the element itself, as it is the product of all elements in the array.

Array with all identical values

How to Handle:

The product will be value^n, which might trigger overflow issues, handled with long or arbitrary precision libraries.

An extremely large array to assess scalability

How to Handle:

Iterative approach with long datatype should work, but consider alternatives if more memory optimization is needed.