Incorrect Arithmetic Slices Solution

Examining code logic

I’m figuring out if the code logic for counting arithmetic slices is correct. It starts by checking if the array has fewer than 3 elements and then iterates to identify slice gaps.

Identifying flaws

The code mishandles arithmetic slices. Despite checking arithmetic properties, it mistakenly re-initializes diff and disregards arithmetic subarrays. This ultimately leads to an incorrect result.

Addressing the error

The code's use of j is correct, as it always trails behind i, comparing nums[i] - nums[j]. This confirms the approach to identifying arithmetic subarray boundaries.

Spotting gaps

I'm thinking through how the logic could skip sub-slices or miscount them. The condition to reinitialize diff when it changes without further validation might lead to incorrect counts.

Mapping out the formula

I’m checking if the standard arithmetic subarray formula n*(n+1)/2 matches the formula (L-2)(L-1)/2. Both expressions should be equivalent for n = i - start - 2.

Adjusting the approach

I'm thinking through whether setting "start = i" might skip arithmetic subarrays ending at "i-1". This adjustment could ensure the code accurately includes such subarrays.

Refining sub-slice handling

I’m determining that the code omits sub-slices ending at "i-1". Adjusting "start = i-1" ensures accurate handling of new differences and partial sub-slices at the end.

Pinpointing potential issues

I’m identifying a bug with "i <= len(nums)" instead of "i < len(nums)" and multiple consecutive difference changes, which might skip arithmetic sub-slices.

Adjusting the start

I’m exploring why setting "start = i" after a difference change might skip sub-slices. The correct continuation should begin at "i-1" to ensure consistent arithmetic progressions.

Understanding boundary errors

The code mismanages boundaries by setting "start = i" post-difference change. This results in skipped slices and fails to handle overlapping arithmetic progressions correctly.

Testing code accuracy

I’m examining the code's handling of arithmetic slices with [1, 2, 3, 5, 7, 9]. It surprisingly returns 2 valid slices, which seems questionable given the sequence.

Pinpointing valid slices

I’m curious about the valid arithmetic slices: [1, 2, 3] and [5, 7, 9]. The middle sub-slice from [3, 5, 7] is skipped due to a difference change.

Pinpointing missing sub-slices

Interestingly, the code misses [3, 5, 7, 9] due to a difference change at index 3, which prevents the accurate identification of valid arithmetic sub-slices beyond this point.

Missed possibilities

The code skips arithmetic sub-slices starting from the middle due to incorrect start adjustments after a difference change. This results in missing potentially valid sequences.

Thought about incorrect solution to Leetcode problem for 35 seconds

First, here’s a quick restatement of what the code is trying to do:

• It keeps track of a “start” index for the current arithmetic run.
• It keeps a running difference (diff) based on the last two elements in the sequence.
• Whenever it detects that the difference has changed, it closes off the old run (computes how many arithmetic sub-slices formed inside it) and starts a new run.

At a glance, the logic looks similar to well-known solutions. However, there is a subtle but critical bug in how it updates the start of the next arithmetic run when it detects a mismatch of differences. This causes the function to miss some valid arithmetic sub-slices.

The Key Bug

Inside the if nums[i] - nums[j] != diff { ... } branch, the code does:

n := i - start - 2
res += n * (n + 1) / 2  // closes out the old arithmetic run
start = i               // starts the new run at i
diff = nums[i] - nums[i-1]

Why is start = i wrong?

When a difference changes at index i, that means the previous run definitely ends at i - 1. But the next arithmetic run might need to begin at i - 1, not i. After all, nums[i-1] and nums[i] form a new pair with a fresh difference that could continue forward.

By setting start = i, you effectively throw away the chance to form an arithmetic sub-slice that begins one step earlier at i-1. As soon as there is a difference change, the code “jumps forward” too far, missing any longer runs that should include the element at i-1.

A Concrete Failing Example

Consider the array:

nums = [1, 2, 3, 5, 7, 9]

Differences across the array:

 (2 - 1) = 1
 (3 - 2) = 1
 (5 - 3) = 2   <-- difference changes here
 (7 - 5) = 2
 (9 - 7) = 2

Let’s see what the code does step-by-step:

1. Indices 0..2: We have an arithmetic run [1,2,3] with difference = 1.

   • At i = 3, it detects nums[3] - nums[2] = 5 - 3 = 2, which is not the old diff=1. So it closes the run.
   • The run length is 3 - 0 = 3. Using the formula n = i - start - 2, we get n = 3 - 0 - 2 = 1. It adds 1 * (1+1)/2 = 1 to res (that correctly accounts for the single 3-length slice [1,2,3]).
   • Then it does start = 3, diff = 5 - 3 = 2.

2. Indices 3..5: Now it looks at [5,7,9] as the new run.

   • It finishes with another run of length 3, adds 1 more arithmetic sub-slice to res.
   • Final result = 2.

But the actual arithmetic sub-slices in [1,2,3,5,7,9] are:

• [1,2,3]
• [5,7,9]
• and [3,5,7,9] (difference = 2 throughout)
• plus the 3-element sub-slice inside it: [3,5,7] is also valid.

The code completely skips counting [3,5,7] and [3,5,7,9] because once it “closed out” the run at index 3, it started fresh at start = 3. It never revisited the possibility that the sub-slice beginning at index 2 (nums[2] = 3) could continue forward with the new difference 2.

In other words, by resetting start = i, the code fails to account for any new arithmetic run that might begin just one step earlier (at i - 1).

Correct Approaches

A proper solution typically does one of two things:

1. Track the length of the current arithmetic run dynamically.

   • If nums[i] - nums[i-1] == nums[i-1] - nums[i-2], increment a count for the “current run” length. Otherwise reset it.
   • Add that count to the result each time because every new element extending the run creates new sub-slices.

2. Use a “start index” but reset it more carefully.

   • When a difference changes at i, you close out the run ending at i - 1.
   • But then you should set start = i - 1 so that the new run starts from that last element.

Either way, you avoid skipping the potential sub-slice that can start just before the mismatch.

Takeaway

Even though the triangular formula for counting sub-slices within a run is correct (n*(n+1)/2 for n = length-2), the place where the code resets start is incorrect. It loses out on valid arithmetic sub-slices that begin one step earlier than i. That’s why it undercounts and is “really incorrect” for certain inputs.