Product of Two Run-Length Encoded Arrays

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Run-length encoding is a compression algorithm that allows for an array to be represented as pairs of (length, value). For example, we can represent the array [1,2,2,3,3,3] with run-length encoding [[1,1],[2,2],[3,3]].

You are given two run-length encoded arrays. Consider arr1 to be the representation of the array nums1 and arr2 to be the representation of the array nums2.

Each array will be run-length encoded in the following format:

The array will start with a pair of positive integers representing the length and value of the first run.
Each pair will be followed by a pair of positive integers representing the length and value of the next run.

For example, arr = [[1, 3], [2, 2]] represents the array [3, 2, 2].

The result of the multiplication of two run-length encoded arrays is the run-length encoded version of the product of the two arrays.

nums1 and nums2 are the arrays represented by arr1 and arr2, respectively. You are to return the run-length encoded version of nums1 * nums2.

Note: Run-length encoding should be done such that the return array strictly follows the format defined above. That is, only runs of the same value should be merged and each run should be represented by a pair of positive integers.

Example 1:

Input: arr1 = [[1,3],[2,2]], arr2 = [[2,1],[3,2],[1,3]]
Output: [[2,3],[3,4]]
Explanation: The first array arr1 represents [3,2,2] and the second array arr2 represents [1,1,2,2,2,3].
The product of them is [3,2,2] * [1,1,2,2,2,3] = [3,2,4,4,4,6].
The run-length encoded version of [3,2,4,4,4,6] is [[1,3],[1,2],[3,4],[1,6]].

Example 2:

Input: arr1 = [[1,3],[2,2]], arr2 = [[1,1],[2,2]]
Output: [[1,3],[3,4]]
Explanation: The first array arr1 represents [3,2,2] and the second array arr2 represents [1,2,2].
The product of them is [3,2,2] * [1,2,2] = [3,4,4].
The run-length encoded version of [3,4,4] is [[1,3],[2,4]].

Constraints:

1 <= arr1.length, arr2.length <= 1000
arr1[i].length == 2
arr2[i].length == 2
1 <= arr1[i][0], arr2[i][0] <= 1000
1 <= arr1[i][1], arr2[i][1] <= 10⁶

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 range of values for the numbers in the run-length encoded arrays, and what data type are they (e.g., integers, floats)? Can the counts be zero?
Can either of the input arrays be empty, or can a run contain a zero count?
If the product results in an empty run-length encoded array, what should the return value be (e.g., an empty array, null, or an array with a single zero value and zero count)?
Are the input arrays guaranteed to be valid run-length encoded arrays (e.g., alternating values and counts, only non-negative counts)?
If consecutive runs in the resulting product have the same value, should I merge them into a single run?

Brute Force Solution

Approach

The brute force method involves fully expanding both encoded sets of numbers into their original, uncompressed forms. Then, we simply multiply corresponding elements to get the product sequence. Finally, this product sequence can be encoded again using run-length encoding to obtain the final answer.

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

First, take the first run-length encoded set of numbers and decompress it. This means turning the compressed representation back into the full, repeated sequence of numbers.
Do the same thing for the second run-length encoded set of numbers, also decompressing it into its full sequence.
Now, take the first number from the first full sequence and multiply it by the first number from the second full sequence. That's the first number in the product sequence.
Continue multiplying corresponding numbers in each full sequence, in order, to create the complete product sequence.
Finally, compress the product sequence using run-length encoding. This means counting consecutive occurrences of the same number and representing them as a number and a repetition count.

Code Implementation

def product_of_run_length_encoded_arrays_brute_force(encoded_array1, encoded_array2):
    expanded_array1 = []
    for value, frequency in encoded_array1:
        expanded_array1.extend([value] * frequency

    expanded_array2 = []
    for value, frequency in encoded_array2:
        expanded_array2.extend([value] * frequency

    product_array = []
    # Need to handle if one array is longer than the other
    for i in range(min(len(expanded_array1), len(expanded_array2))):
        product_array.append(expanded_array1[i] * expanded_array2[i])

    if not product_array:
        return []

    run_length_encoded_product_array = []
    current_value = product_array[0]
    current_frequency = 1

    # Need to compress the results before returning
    for i in range(1, len(product_array)):
        if product_array[i] == current_value:
            current_frequency += 1
        else:
            run_length_encoded_product_array.append([current_value, current_frequency])
            current_value = product_array[i]
            current_frequency = 1

    # Important to add the last value and count pair after loop
    run_length_encoded_product_array.append([current_value, current_frequency])

    return run_length_encoded_product_array

Big(O) Analysis

Time Complexity

O(N) – The dominant operations are decompressing the two run-length encoded arrays into their full representation and then iterating through these decompressed arrays to compute the product. Let N be the length of the final decompressed array, assuming both decompressed arrays have lengths proportional to N. Decompressing each run-length encoded array takes time proportional to the final length of the array after decompression, so this contributes O(N). Multiplying corresponding elements from the two decompressed arrays also takes O(N) time. The final run-length encoding of the product sequence is also O(N). Therefore the overall time complexity is O(N).

Space Complexity

O(N) – The brute force method first decompresses both run-length encoded arrays into their original forms. If the total length of the decompressed arrays is N (where N is the sum of lengths after expansion of both arrays), auxiliary space of O(N) is required to store these two expanded arrays. The product sequence, also of length N in the worst case, is created and requires additional O(N) space. The final run-length encoded representation of this product sequence could potentially be of size N in the worst case, so further expansion requires O(N). Therefore, the overall space complexity is O(N).

Optimal Solution

Approach

The most efficient way to solve this involves processing both encoded lists simultaneously. We cleverly track how much of each encoded element we've 'consumed' to compute the product without fully expanding either list. By advancing through both encoded lists strategically, we calculate the product on-the-fly and only when both lists contribute to the result.

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

Imagine you have two streams of numbers, each compressed. Each compressed stream consists of pairs: a number and how many times it repeats.
We'll walk through both streams at the same time. We look at the current repetition count for each stream.
We figure out the minimum of the two repetition counts. This tells us how many times we can multiply the current numbers from both streams.
Multiply the two current numbers, and then multiply the result by the minimum repetition count we just calculated.
Add this result to a running total, which will be our final product.
Now, we reduce the repetition counts of both streams by the minimum we found. The stream with the smaller count will have its repetition count reduced to zero.
If a stream's repetition count reaches zero, we move to the next number and repetition count in that stream.
We continue this process until we've processed all the numbers and repetition counts in both streams.
The final running total is the product of the two run-length encoded arrays.

Code Implementation

def product_of_run_length_encoded_arrays(encoded_array1, encoded_array2):
    index_array1 = 0
    index_array2 = 0
    product = 0

    while index_array1 < len(encoded_array1) and index_array2 < len(encoded_array2):
        value1, frequency1 = encoded_array1[index_array1]
        value2, frequency2 = encoded_array2[index_array2]

        # Determine the minimum frequency
        minimum_frequency = min(frequency1, frequency2)

        # Calculate and add the product
        product += value1 * value2 * minimum_frequency

        # Update frequencies
        encoded_array1[index_array1] = (value1, frequency1 - minimum_frequency)
        encoded_array2[index_array2] = (value2, frequency2 - minimum_frequency)

        # Move to the next element in array1 if frequency is zero
        if encoded_array1[index_array1][1] == 0:
            index_array1 += 1

        # Move to the next element in array2 if frequency is zero
        if encoded_array2[index_array2][1] == 0:
            index_array2 += 1

    return product

Big(O) Analysis

Time Complexity

O(m + n) – The algorithm iterates through two run-length encoded arrays simultaneously, where 'm' is the number of pairs in the first array and 'n' is the number of pairs in the second array. In each step, we compare the repetition counts of the current elements in both arrays and advance the pointer of the array with the smaller count. We process each pair in both arrays at most once. Therefore, the time complexity is proportional to the sum of the number of pairs in both arrays, resulting in O(m + n).

Space Complexity

O(1) – The algorithm uses a constant amount of extra space. It involves tracking the current index in each run-length encoded array, along with variables to store the current values and repetition counts being processed. The amount of space needed for these variables does not depend on the size of the input arrays; it remains constant regardless of the number of elements in the encoded arrays. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
One or both input arrays are null or empty	Return an empty list or throw an IllegalArgumentException, depending on desired behavior, as there's no product to compute.
Run-length encoded arrays with zero lengths	The product for a length of zero should be zero, handled by multiplying the value by the decoded length during calculation.
Large lengths in run-length encoded arrays, leading to potential integer overflow in the product	Use long data type for storing the product to prevent integer overflow, or return appropriate error message if overflow is unavoidable.
Values in the arrays are zero	Any product involving a zero value should be zero; account for this during the product calculation.
Very large decoded arrays, exceeding memory capacity	Process arrays lazily or in chunks to avoid loading the entire decoded array into memory, implementing some form of streaming approach.
Arrays with only one run (single element)	The solution should correctly handle arrays with a single run by multiplying the value by its length.
Consecutive runs in one or both arrays result in zero values after multiplication	The solution should correctly generate the result and run length when the product of values equals zero, potentially resulting in a sequence of [0, length].
The final result array has consecutive identical products; the solution can combine them to be run-length encoded	Iterate through the product array and merge any adjacent runs with identical product values.

One or both input arrays are null or empty

How to Handle:

Return an empty list or throw an IllegalArgumentException, depending on desired behavior, as there's no product to compute.

Run-length encoded arrays with zero lengths

How to Handle:

The product for a length of zero should be zero, handled by multiplying the value by the decoded length during calculation.

Large lengths in run-length encoded arrays, leading to potential integer overflow in the product

How to Handle:

Use long data type for storing the product to prevent integer overflow, or return appropriate error message if overflow is unavoidable.

Values in the arrays are zero

How to Handle:

Any product involving a zero value should be zero; account for this during the product calculation.

Very large decoded arrays, exceeding memory capacity

How to Handle:

Process arrays lazily or in chunks to avoid loading the entire decoded array into memory, implementing some form of streaming approach.

Arrays with only one run (single element)

How to Handle:

The solution should correctly handle arrays with a single run by multiplying the value by its length.

Consecutive runs in one or both arrays result in zero values after multiplication

How to Handle:

The solution should correctly generate the result and run length when the product of values equals zero, potentially resulting in a sequence of [0, length].

The final result array has consecutive identical products; the solution can combine them to be run-length encoded

How to Handle:

Iterate through the product array and merge any adjacent runs with identical product values.