Problem Description

The given problem presents a run-length encoded list, which is a common method for compressing data in which consecutive elements are replaced by just one value and the count of that value. The nums list contains pairs where the first element of each pair represents how many times the next element should appear in the decompressed list.

For instance, if we have a compressed list [3, 4, 1, 2], it means we have 3 of 4s and 1 of 2 in the decompressed list: [4, 4, 4, 2].

The challenge is to convert this run-length encoded list into a decompressed list where each pair [freq, val] is expanded into freq instances of val.

Intuition

To solve this problem, we iterate over the given list nums to process each [freq, val] pair. Because freq and val are always adjacent, starting with freq, we skip every other element by using a loop that begins at index 1 and moves in steps of 2. This allows us to directly access each val while its corresponding freq is just the previous element.

For each freq and val pair, we replicate val, freq times. The replication can be efficiently accomplished in Python with list multiplication ([val] * freq). This creates a list of val repeated freq times. We then extend our result list with this new list.

The solution is achieved in a straightforward manner through this looping and extending, transforming the encoded list into the desired decompressed version in line with the problem requirements.

Solution Approach

The provided solution employs a simple yet efficient method that directly corresponds to the idea of run-length decoding. We use a basic for loop to iterate over the input list nums. The list data structure is the only structure used, which is apt for this task, as Python lists provide an efficient way to manage sequence data and support operations required for the solution.

Here's a step-by-step breakdown of the solution:

1. We initialize an empty list res which will hold our decompressed elements.

2. The for loop starts at index 1 and iterates till the end of the nums list with a step of 2. This step is important because our freq is always the element just before val (i.e., at index i - 1), and val is at every second index starting from 1 (which is i in this context).

3. In each iteration, we create a list with val repeated freq times. This is done by multiplying a single-element list containing val by freq ([nums[i]] * nums[i - 1]).

4. We then extend the res list with this repeated list. The .extend() method is used because it adds elements of the list argument to the end of the current list, effectively concatenating them to res.

5. Once the loop has processed all the [freq, val] pairs, the res list contains the fully decompressed list, which is then returned.

Through the use of list indexing, list multiplication, and the .extend() method, the provided Python solution efficiently decompresses the run-length encoded list with minimal complexity and optimal use of Python's list capabilities.

Here is the code snippet embedded in markdown for clarity:

class Solution:
    def decompressRLElist(self, nums: List[int]) -> List[int]:
        res = []
        for i in range(1, len(nums), 2):
            res.extend([nums[i]] * nums[i - 1])
        return res

This approach is linear in time complexity (O(n)), where n is the number of elements in the input list nums, as it requires a single pass through the list. The space complexity is also linear (O(n)), as it directly depends on the total number of elements that will be in the decompressed list.

Example Walkthrough

Let's take a small example to illustrate the solution approach. Suppose we are given the following compressed list:

[2, 5, 3, 6]

This implies we should have 2 of ‘5’s and 3 of ‘6’s in our decompressed list.

Following the steps of the solution:

1. Initialize an empty list res. This will store the final decompressed list.

2. Begin a loop starting at index 1 (i = 1), since it's the position of the first value to be repeated. We will increment the index by 2 every iteration to jump to the next value to be repeated.

3. At index i = 1, the element is 5, and the frequency (freq) is the previous element (nums[i - 1]), which is 2. We create a list with 5 repeated 2 times: [5, 5].

4. Extend the res list with this new list: res now becomes [5, 5].

5. Move to the next value in the list that needs to be repeated, which is at index i = 3 (the element is 6). The frequency for 6 is 3 (nums[i - 1]).

6. Repeat the value 6, 3 times to get [6, 6, 6].

7. Extend the res list with this new list: res becomes [5, 5, 6, 6, 6].

After the loop has finished, the res list contains the decompressed list that we need, and we return it:

[5, 5, 6, 6, 6]

This approach directly transforms the encoded list [2, 5, 3, 6] into the desired decompressed version [5, 5, 6, 6, 6].

The given Python code, when executed with our example nums, would return [5, 5, 6, 6, 6] as expected:

class Solution:
    def decompressRLElist(self, nums: List[int]) -> List[int]:
        res = []
        for i in range(1, len(nums), 2):
            res.extend([nums[i]] * nums[i - 1])
        return res

The provided example demonstrates how the solution method applies to a specific instance, reflecting the simplicity and effectiveness of the approach for decompressing a run-length encoded list.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity is O(n), where n is the total number of elements in the input list nums. This is because the loop iterates through every other element of nums, performing a constant amount of work for each pair of freq-value (since the .extend() method is called once per pair), and the number of pairs is n/2.

Space Complexity

The space complexity is also O(n), but here n represents the total number of elements in the decompressed list res. Although the input list size is halved due to the pairs, the decompression may lead to a larger output. In the worst case, all the freq-values are large, causing the output list size to be significantly larger than the size of the input list.

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