Question textA more-challenging functionSequences of numbers in which there are frequent "runs" of a particular number repeating several times can often be more compactly represented by use of what is called "run length encoding". A list is run-length encoded by representing it as a list of pairs (2-tuples), where each pair is a number and the length of the "run" of that number, where the length is 1 if a number occurs once, 2 if it occurs twice in a row, etc. Write a function run_length_encode(nums) that returns the run-length encoded representation of the list of integers, nums.For example:[table]
Test | Result data = [5, 5, 5, 10, 10] print(run_length_encode(data)) | [(5, 3), (10, 2)] data = [10, 20, 30, 30, 30, 30] print(run_length_encode(data)) | [(10, 1), (20, 1), (30, 4)]
[/table] Answer:(penalty regime: 0, 10, 20, ... %)

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BlackTom AI · Accepted Answer

The prompt describes run-length encoding: given a list of integers, return a list of 2-tuples where each tuple contains a value and the number of times that value repeats consecutively.

Since there are no answer options to critique, I will outline a clear, step-by-step approach to implement run_length_encode(nums) and explain the rationale behind it, including handling edge cases and correctness considerations.

Step 1: Understand the input and expected output
- Input: nums, a list of integers (could be empty).
- Output: a list of tuples [(value1, run_length1), (value2, run_length2), ...] where each run_lengthi is the length of the consecutive repetition of valuei in nums.
- Example checks:
  - [5, 5, 5, 10, 10] -> [(5, 3), (10, 2)]
  - [10, 20, 30, 30, 30, 30] -> [(10, 1), (20, 1), (30, 4)]

Step 2: Decide on a plan to traverse the list
- Initialize an empty result list res.
- If nums is empty, return an empty list.
- Start with the first element as the current value (cur = nums[0]) and a count (cnt = 1).
- Iterate through the rest of nums, and for each element x:
  - If x == cur, increment cnt (we’ve extended the current run).
  - Otherwise, append (cur, cnt) to res, then set cur = x and reset cnt = 1 (start a new run).
- After the loop ends, append the final (cur, cnt) to res to capture the last run.

Step 3: Consider time and space complexity
- Time complexity: O(n), where n is the length of nums, since we traverse the list once.
- Space complexity: O(k), where k is the number of runs (in the worst case k = n if no consecutive repeats).

Step 4: Handle edge cases
- Empty input: return [] without error.
- Single-element list: return [(nums[0], 1)].
- Negative numbers or large integers: no special handling needed beyond treating values as keys for equality checks.

Step 5: Provide a concrete implementation (descriptive, not using code blocks)
- If nums is empty, return [].
- Set cur to the first element and cnt to 1.
- For each subsequent element x in nums:
  - If x equals cur, increase cnt by 1.
  - Otherwise, append (cur, cnt) to the result, then set cur = x and cnt = 1.
- After processing all elements, append the final pair (cur, cnt) to the result.
- Return the result list.

Optional: Quick Python sketch (readable inline, not a block):
- def run_length_encode(nums):
-     if not nums:
-         return []
-     res = []
-     cur, cnt = nums[0], 1
-     for x in nums[1:]:
-         if x == cur:
-             cnt += 1
-         else:
-             res.append((cur, cnt))
-             cur, cnt = x, 1
-     res.append((cur, cnt))
-     return res

Step 6: Validate with the given examples mentally
- For [5, 5, 5, 10, 10], we start with (5, 3), then (10, 2), yielding [(5, 3), (10, 2)].
- For [10, 20, 30, 30, 30, 30], we get (10, 1), (20, 1), (30, 4), yielding [(10, 1), (20, 1), (30, 4)].

In summary, the method above faithfully captures the idea of run-length encoding by emitting a new tuple whenever a run ends and tallying the length of each run as we go, which aligns with the standard definition and the provided examples.

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