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- bahii honii we have “Count the Number of Incremovable Subarrays I”

—> bahiii honii we have 0-indexed array of postive intergers “nums” 

- A Subarray of “nums” is called “incremovable” if nums becomes strictly increasing on removing the subarray
- bahii ke exmple : the subarray [3,4] , is an incremovable subarray of [5,3,4,6,7] because removing this subarray changes the array [5,3,4,6,7] to [5,6,7] which is striclty increasing
- bahiii honii we need to return thr total number of “incremovable” subarrays of nums
- bahii awll 7aje “ an empty array is considered striclty increasing”

→ bahii the “subarray” is a contigous non-empty sequance of elements within an array 

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Q- what is the input output & format ?

- input : List[int]
- output : int

Q- can you go through an exmple ? 

nums=[1,2,3,4] 

bahii awll 7aje we need to check if all the subarray and how to removing them will work yes or non ? 

Q- let’s try to check the naive solution here ? 

a naive solution will be just generate all the possible subarray 

then remove them( vittualy ) and chekc if the remaining array is sorted okii yes or non ? 

generate for me all the subarray of this  [1,2,3,4]

Sure! Here are all the subarrays of the array `[1, 2, 3, 4]`:

1. `[1]`
2. `[2]`
3. `[3]`
4. `[4]`
5. `[1, 2]`
6. `[2, 3]`
7. `[3, 4]`
8. `[1, 2, 3]`
9. `[2, 3, 4]`
10. `[1, 2, 3, 4]`

These are all the possible contiguous subarrays of the given array.

next for each one remove the subarray form the orginal array and return it 

generete all the reaming arrr after removing the subarray from the last generated ones 

Certainly! Let's generate the remaining arrays after removing each subarray from the original array `[1, 2, 3, 4]`.

1. Removing `[1]`:
    - Remaining array: `[2, 3, 4]`
2. Removing `[2]`:
    - Remaining array: `[1, 3, 4]`
3. Removing `[3]`:
    - Remaining array: `[1, 2, 4]`
4. Removing `[4]`:
    - Remaining array: `[1, 2, 3]`
5. Removing `[1, 2]`:
    - Remaining array: `[3, 4]`
6. Removing `[2, 3]`:
    - Remaining array: `[1, 4]`
7. Removing `[3, 4]`:
    - Remaining array: `[1, 2]`
8. Removing `[1, 2, 3]`:
    - Remaining array: `[4]`
9. Removing `[2, 3, 4]`:
    - Remaining array: `[1]`
10. Removing `[1, 2, 3, 4]`:
    - Remaining array: `[]`

These are the remaining arrays after removing each subarray from the original array `[1, 2, 3, 4]`.

Q- as we can see all the remaining arrays are increasing in there element so we just need to keep a counter to track them okii 

Q- i think the intution is clear here but the challange is how to translate this mental model idea into an actual implementation 

Q- is the intuion in the right direction yes or non 

Q- what are the conrete challanges here ? 

—> one is how to efficently “scan” all the subarrays 

—> then how to to check or generate the remaining element in the array without the subarray 

Q- okii let’s try to build a mental model to tackle the first challange ? 

- the first one is to tackle the scaning of all the subarrays

one why given that they are a contigous block of element we need to use slide window technique right ? which redcuing the scaning form O(n**2) worse case scenario into O(n)  one 

- technically to implment this we need a kind to two pointer okii (l and r) that scan the whole array

Q- for exmple this pseudoCode ? 

```python
l=0
n=len(nums)
for r in range(n):
		current_sub_arr=nums[l:r]
		
```

Q- did my thinking process so far in the right direction yes or non ? 

Q- without given me the full solution oki

Yes, your thinking process is in the right direction.  You've correctly identified the core challenges and proposed reasonable approaches:

- **Naive approach understanding:** You understand the brute-force method and its limitations.
- **Incremovability check:** You recognize that checking for strict increasing order is a key part of the solution.
- **Subarray generation (Challenge 1):** You've correctly identified the inefficiency of generating all subarrays explicitly and proposed sliding window (two pointers l and r) as a way to improve efficiency. Your pseudocode snippet for iterating with l and r is a good start.
- **Remaining array generation (Challenge 2):** You've pointed out the need to efficiently generate/check the remaining array after subarray removal. This is where further optimization and cleverness will come in. You haven't solved this part yet, but recognizing the challenge is a crucial step.

Next steps you should consider:

1. **How to check for "strictly increasing" efficiently:** Think about how you can determine if a subarray (or the remaining elements) is strictly increasing *without* creating a new array each time. Can you do it by directly comparing elements within the original nums array using your l and r pointers?
2. **Refining the sliding window:** Your initial l and r loop is a good start, but how will you handle the *removal* of the subarray conceptually? The key is to not actually remove elements but rather to think about which parts of nums you are *considering* as the "remaining array" based on the current positions of l and r.
3. **Counting incremovable subarrays:** Think about when you should increment your count of incremovable subarrays based on the checks you perform. This will tie into how you define your l and r movements.

You're on the right track!  Keep thinking about how to efficiently determine the increasing property and how to virtually "remove" the subarray within your sliding window logic.  Don't give up!

Q- tack the first question : 

1. **How to check for "strictly increasing" efficiently:** Think about how you can determine if a subarray (or the remaining elements) is strictly increasing *without* creating a new array each time. Can you do it by directly comparing elements within the original nums array using your l and r pointers?

"—> We need to find a technique that lets us check the remaining elements without creating a new array each time.

—> So, we can leverage the left and right pointers instead.

remaining_nums = nums[
] + nums[r+1:]

—> Could you provide a manual proof of this approach by working through an example?"

This idea suggests using pointers to define segments of the array rather than actually creating new arrays, which can be more efficient. Let me know if you'd like help with a manual proof or example for this method!

You're absolutely right about leveraging the l and r pointers and *not* creating new arrays.  Your idea of remaining_nums = nums[:l] + nums[r+1:] is the correct way to *conceptually* represent the remaining elements.  However, you don't need to create this new remaining_nums list explicitly.

Here's how you can manually "execute" this concept to check for strict increasing without creating new arrays:

**Example nums = [5, 3, 4, 6, 7]**

Let's say your l pointer is at 1 and your r pointer is at 2. This means the current subarray you're considering removing is nums[l:r+1] = nums[1:3] = [3, 4].

1. **Check the left portion:** You need to check if nums[:l] (which is nums[:1] or [5] in this case) is strictly increasing. Since it's just one element, it is considered strictly increasing.
2. **Check the right portion:** You need to check if nums[r+1:] (which is nums[3:] or [6, 7] in this case) is strictly increasing. You can do this by iterating from i = r+1 to the end of nums and comparing nums[i] with nums[i-1] (for i > r+1). In this case, 7 > 6, so the right portion is strictly increasing.
3. **Check across the "gap":** Finally, and this is crucial, you must check across the "gap" created by the removed subarray. You need to compare the last element of the left portion (if it exists) with the first element of the right portion (if it exists). In our example:
    - Last element of the left: nums[l-1] which is nums[0] or 5.
    - First element of the right: nums[r+1] which is nums[3] or 6.
    - Is nums[r+1] > nums[l-1]? Yes, 6 > 5.

Since all three conditions are met (left increasing, right increasing, and across the gap increasing), you would increment your count of incremovable subarrays for this specific l and r combination.

**Key point:** You are doing all these checks *directly on the original nums array* using index comparisons.  You never have to create and store new subarrays.  This is how you make your solution efficient.

Now, think about how you would incorporate this logic into your nested l and r loops to iterate through all possible subarrays and count the incremovable ones.

---

Q- okii let’s try to translate the progess so far ? 

```python
l=0
n=len(nums)
for r in range(n):
	// check the removable logic 
	// gap 
	if nums[r+1]>nums[l]:
		
```