Problem Description

The problem provides us with a compression technique called "run-length encoding". This technique compresses an array of integers by representing segments of consecutive repeated numbers as a 2D array. Each entry in the encoded array has two elements: the value being repeated and the frequency of that value.

For example, nums = [1,1,1,2,2,2,2,2] is encoded as encoded = [[1,3],[2,5]], meaning that the number 1 is repeated 3 times followed by number 2 which is repeated 5 times.

The task is to calculate the product of two run-length encoded arrays, encoded1 and encoded2. To find the product, we need to:

1. Expand encoded1 and encoded2 back into the original full arrays (nums1 and nums2).
2. Multiply the corresponding elements of nums1 and nums2 to form a new array prodNums.
3. Compress prodNums back into run-length encoded format.

We need to ensure that the final encoded product array is as short as possible.

Intuition

The intuition behind the solution is to simulate the product of the two arrays without fully expanding them. This is important for efficiency, especially when the encoded arrays represent very long sequences.

By iterating through the segments of the first array, we:

• Keep track of how many times we have used the value in the current segment.
• Determine the product with the corresponding part of the second encoded array.
• Update the result by either adding a new segment or extending the last segment if the product value is the same.
• Keep track of the portion of the segment of the second array that has been consumed and move to the next segment once we have used it up.

We use the minimum frequency from the current segments of encoded1 and encoded2 to decide how to extend or create the new segments in the resulting array. This way, we simulate the expansion and multiplication steps by directly compressing the product. We avoid unnecessary computation and achieve the efficient calculation of the run-length encoded product.

Learn more about Two Pointers patterns.

Solution Approach

The implementation of the solution is straightforward in logic but requires careful handling of the indices and frequencies from the encoded arrays. Here's a step-by-step breakdown of how the solution works:

1. Initialize an empty list ans, which will store our result in run-length encoded format.
2. Initialize a variable j to keep track of our position in encoded2 while we iterate through encoded1.
3. Loop through each segment [val_i, freq_i] of encoded1. For each segment, we do the following:
   • While freq_i is greater than zero:
     • We take the minimum of freq_i and the frequency of the current segment in encoded2 (i.e., encoded2[j][1]). This decides how much of the segment we can use in this step of the product.
     • Calculate the product v of val_i and the value of the current segment in encoded2 (i.e., encoded2[j][0]).
     • Check if the last segment in ans has the same value as v. If so, increase the frequency of that segment by the frequency we are currently using f. This step ensures we are compressing the result as we go.
     • If the last segment in ans has a different value, append a new segment [v, f] to ans.
     • Decrease freq_i by f to keep track of the remaining frequency that needs to be accounted for from the current segment in encoded1.
     • Decrease the frequency of the current segment in encoded2 by f as we have used up f frequency of it.
     • If the current segment in encoded2 is fully used (frequency becomes zero), move to the next segment in encoded2 by incrementing j.

By following these steps, the while loop effectively takes care of creating the product without ever needing to fully expand the encoded arrays. Once we've finished processing all segments in encoded1, we're left with ans which contains our compressed product.

This approach uses a single pass and keeps space complexity low, as we never create the fully expanded arrays. It combines elements of two-pointers technique—i iterating over encoded1 and j keeping track of our place in encoded2—with a merge-like operation where we merge contributions from corresponding segments.

Example Walkthrough

Let's illustrate the solution approach with a small example. Suppose we have two run-length encoded arrays:

encoded1 = [[2,3],[3,2]] and encoded2 = [[6,1],[3,3],[9,1]]

The first array encoded1 can be expanded to [2,2,2,3,3] and the second array encoded2 to [6,3,3,3,9]. We are to find the product of these two arrays and then compress it back to run-length encoding format. Let's follow the steps in the solution approach.

1. Initialize ans = [], an empty list to hold the result.

2. Initialize j = 0, a variable to keep track of our position in encoded2.

3. Begin looping over encoded1. Our first segment is [2,3] indicating three 2s.

   • While freq_i (3 for the first segment of encoded1) is greater than zero:
     • The first segment of encoded2 has only one element left, so we take the minimum of freq_i and encoded2[j][1], which is min(3, 1) = 1.
     • The product v of values 2 (val_i) and 6 (encoded2[j][0]) is 12.
     • ans = [] is currently empty, so we append [12, 1] to ans.
     • We decrease freq_i by 1, now freq_i is 2.
     • We also decrease encoded2[j][1] by 1, and since it would now be 0, we move to the next segment by incrementing j (now j=1 and segment [3,3] of encoded2 is the current).

4. We continue the while loop because freq_i is still greater than zero:

   • Now we have freq_i = 2 and the frequency of the current segment in encoded2[j][1] is 3.
   • We take the minimum, which is 2.
   • The next product v of values 2 and 3 is 6.
   • ans = [[12, 1]], and as 6 is different from 12, we append [6, 2] to ans.
   • Both freq_i and encoded2[j][1] are decreased by 2, resulting in freq_i = 0 and encoded2[j][1] = 1.

5. Since freq_i is zero, we move to the next segment of encoded1.

6. The next segment is [3,2]. We repeat a similar process here:

   • We have freq_i = 2 and encoded2[j][1] is 1 (from the remaining part of the previous segment [3,1]).
   • We take the minimum, which is 1.
   • The product of values 3 and 3 is 9.
   • Since the last value in ans is 6, we add a new segment [9, 1] to ans.
   • Decrease freq_i to 1 and since encoded2[j][1] is now 0, we increase j to 2 (next segment [9,1] of encoded2).

7. We continue in the loop with the remaining freq_i = 1 for value 3 in encoded1:

   • We have freq_i = 1 and encoded2[j][1] is also 1.
   • The product of values 3 and 9 is 27.
   • As 27 is different from the last value in ans, we add another segment [27, 1] to ans.
   • Both freq_i and encoded2[j][1] are decreased to 0.

Finally, we're finished processing all segments in encoded1 and encoded2, resulting in ans = [[12, 1], [6, 2], [9, 1], [27, 1]] which is the run-length encoded product of nums1 and nums2.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code primarily depends on these factors:

1. Iterating through encoded1, which has n elements: O(n).
2. A nested loop for processing elements in encoded2, which can be iterated up to m times in the worst case (in case the sizes in encoded1 are much larger): O(m).

However, since the second loop decreases the frequency from encoded2 without resetting, each element of encoded2 will be processed at most once throughout the entire iteration of encoded1. Therefore, the total time complexity will be O(n + m) where n is the total number of elements in encoded1 and m is the total number of elements in encoded2.

Space Complexity

The space complexity is determined by the size of the output, which in the worst case might contain an individual element for each multiplication of pairs from encoded1 and encoded2. In the worst case, every pair multiplication might result in a distinct value not matching the last element of the ans list, thus:

1. The ans list: Up to O(n + m) in the worst case.
2. Constant space for the pointers and temporary variables like vi, fi, f, v, j.

Therefore, the total space complexity of the given code would be O(n + m).

Learn more about how to find time and space complexity quickly using problem constraints.