Solution Explanation for 1868. Product of Two Run-Length Encoded Arrays

This problem involves manipulating run-length encoded arrays. The core idea is to efficiently compute the product of two arrays represented in this compressed format without explicitly expanding them to their full lengths.

Algorithm:

The solution iterates through encoded1. For each segment in encoded1, it determines how much of that segment overlaps with the current segment in encoded2. It calculates the product of the values and the overlapping frequency. This product and frequency form a new segment in the result. If the new segment's value matches the last segment in the result, the frequencies are merged. Otherwise, a new segment is appended to the result.

Time Complexity:

The algorithm iterates through each segment in encoded1 once. The inner while loop iterates at most as many times as the total number of elements in the expanded arrays. In the worst-case scenario, where there's minimal compression, this could be O(m*n), where 'm' and 'n' are the number of elements in the expanded versions of encoded1 and encoded2, respectively. However, given the constraints, it is realistically bounded by O(max(M,N)) where M and N are the lengths of encoded1 and encoded2 respectively. Appending and updating segments in the result array takes constant time per operation. Thus, the overall time complexity is O(M+N).

Space Complexity:

The space complexity is determined by the size of the resulting run-length encoded array ans. In the worst case, the output array could have a length equal to the length of the longer of the two input arrays; therefore, the space complexity is O(max(M,N)). This is due to the fact that the maximum size of the output array is linearly dependent on the maximum size of the input arrays.

Code Explanation (Python):

class Solution:
    def findRLEArray(
        self, encoded1: List[List[int]], encoded2: List[List[int]]
    ) -> List[List[int]]:
        ans = []
        j = 0
        for vi, fi in encoded1:  # Iterate through segments of encoded1
            while fi:  # Iterate while frequency of current segment > 0
                f = min(fi, encoded2[j][1])  # Overlapping frequency
                v = vi * encoded2[j][0]  # Product of values
                if ans and ans[-1][0] == v:  # Merge if same value
                    ans[-1][1] += f
                else:
                    ans.append([v, f])  # Append new segment
                fi -= f  # Adjust remaining frequency in encoded1
                encoded2[j][1] -= f  # Adjust remaining frequency in encoded2
                if encoded2[j][1] == 0:  # Move to next segment in encoded2
                    j += 1
        return ans

The code in other languages (Java, C++, Go) follows the same logic and has equivalent time and space complexity. The key operations are the nested loops iterating through segments and the comparisons/merges to ensure minimal output length.