680. Valid Palindrome II
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Difficulty: Easy
Topics: String, Two Pointers
Companies:
Problem Statement:
Given a string s, return true if the s can be palindrome after deleting at most one character from it.
Example 1:
     Input: s = "aba"
Output: true
   
content_copy Use code with caution.
Example 2:
     Input: s = "abca"
Output: true
Explanation: You could delete the character 'c'.
   
content_copy Use code with caution.
Example 3:
     Input: s = "abc"
Output: false
   
content_copy Use code with caution.
Constraints:
1 <= s.length <= 10<sup>5</sup>
s consists of lowercase English letters.

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Clear problem title and difficulty level.
Relevant topics and company tags (if available).
Detailed problem description with clear examples.
Constraints for the input.


Q- can you explain the problem using your own words ?

Oky so we are a string , we need to check is this string can be a palindrome even when we delete at most one character oky .. 
Q- what is the definition of a valid palindrom ? 
A sequence of characters that read forwards and backwards the same manner 

Q- what is the input/output format? 

The input: str 
The output : str 

Q- can go through an example? 

“Abca” 
Going in both directions 
a ==a 
Move b!=c 
We remove c 
We get a “b” 
So “aba” is the same here okay is a valid palindrome 

Q- can you see any pattern or technique to solve it? 

I think of using two pointers to traverse the sequence forward and backward 
Left =0
Right = len(s)-1
While left < right 

But also this only checks the this string after removing only one characters 

Oky we implement check palindrome as helper function using two pointer technique 
For the deletion part we use a greedy approach where we delete one of first two couples that does not match 
I mean two couples because we will use alos two pointer technique to traverse both forward and backward .. for the first two element that does not match we just remove one oky 
Concrete example 


“Abca” 
A ->left == a <-right 
Move the two-pointers 
b → left == c ← right 
Remove the left one “b” (you can also remove the right one c)

	Now we break for the loop 
And make a call to check_plaindrome function and return it’s result 


Q- before i continue use mathematics proof and hypthotics chekking 
To accept or denied my thinking proess here oky 

Here’s a formatted version of your content:

---

### You're on the right track with your thought process! Let's go through it step by step and use some mathematical reasoning to confirm whether your approach is valid.

---

### Understanding the Problem:

- **Goal**: Determine if the given string `s` can be a palindrome after deleting at most one character.
- **Palindrome Definition**: A string that reads the same forward and backward.

---

### Your Proposed Solution:

#### 1. Two-Pointer Technique:
- Use two pointers: `left` starting at the beginning of the string, and `right` starting at the end.
- Move `left` and `right` towards each other while comparing the characters at these positions.

#### 2. Handling Mismatches:
- When a mismatch is found (`s[left] != s[right]`), attempt to skip either the character at `left` or `right`.
- Check if the resulting substrings are palindromes.

#### 3. Helper Function:
- Implement a helper function to check if a substring is a palindrome using the two-pointer technique.

---

### Mathematical Justification:

#### Why Skipping One Character Works:
- If the original string is not a palindrome, it could be due to at most one mismatched pair.
- By removing one character (either at `left` or `right`), we might eliminate the cause of the mismatch.
- This works because palindromes are symmetrical, so the mismatch must be around the center.

#### Ensuring All Cases Are Covered:
- By checking both possibilities (removing `s[left]` or `s[right]`), we account for cases where the mismatch could be resolved by removing either character.

---

### Testing Your Approach with Examples:

#### 1. Example: `"abca"`

```
Index:   0  1  2  3
String: 'a''b''c''a'
```

- **Step 1**: `left = 0`, `right = 3`
  - `s[0] == s[3] ('a' == 'a')` ➔ Move `left` to 1, `right` to 2
- **Step 2**: `left = 1`, `right = 2`
  - `s[1] != s[2] ('b' != 'c')` ➔ Mismatch found
- **Attempt to Skip a Character**:
  - **Option 1**: Skip `s[1] ('b')`
    - Check if `s[2:3] ('c''a')` is a palindrome
    - `s[2] != s[3]` ➔ Not a palindrome
  - **Option 2**: Skip `s[2] ('c')`
    - Check if `s[1:2] ('b')` is a palindrome
    - Single character ➔ Always a palindrome
  - **Conclusion**: Since one of the options results in a palindrome, return `True`.

#### 2. Example: `"abc"`

- **Step 1**: `left = 0`, `right = 2`
  - `s[0] != s[2] ('a' != 'c')` ➔ Mismatch found
- **Attempt to Skip a Character**:
  - **Option 1**: Skip `s[0] ('a')`
    - Check if `s[1:3] ('b''c')` is a palindrome
    - `s[1] != s[2]` ➔ Not a palindrome
  - **Option 2**: Skip `s[2] ('c')`
    - Check if `s[0:2] ('a''b')` is a palindrome
    - `s[0] != s[1]` ➔ Not a palindrome
  - **Conclusion**: Neither option results in a palindrome, return `False`.

#### 3. Edge Case: `"deeee"`

- **Step 1**: `left = 0`, `right = 4`
  - `s[0] != s[4] ('d' != 'e')` ➔ Mismatch found
- **Attempt to Skip a Character**:
  - **Option 1**: Skip `s[0] ('d')`
    - Check if `s[1:5] ('e''e''e''e')` is a palindrome ➔ Palindrome confirmed
  - **Option 2**: Skip `s[4] ('e')`
    - Check if `s[0:4] ('d''e''e''e')` is not a palindrome
  - **Conclusion**: Since skipping `s[0]` results in a palindrome, return `True`.

---

### Mathematical Proof of Correctness:

#### Completeness:
- By checking both possibilities when a mismatch occurs, you consider all cases where removing one character could fix the palindrome.
- This ensures that if it's possible to form a palindrome by deleting at most one character, your algorithm will find it.

#### Efficiency:
- The two-pointer approach operates in `O(n)` time.
- The helper function checks substrings in `O(n)` time, but since it only runs when a mismatch is found (and only twice at most), the overall time complexity remains `O(n)`.

---

### Possible Edge Cases to Consider:
- **Strings with Repeating Characters**: Ensure your algorithm handles cases like `"aaaaabaaaaa"`.
- **Strings of Length 1 or 2**: Single characters are palindromes by default. For strings of length 2, check if removing one character results in a palindrome.
- **Case Sensitivity and Non-Alphanumeric Characters**: According to the problem constraints, you only need to handle lowercase English letters, so no additional processing is needed.

---

### Conclusion:

Your approach is mathematically sound and efficiently addresses the problem. You correctly:

- Use the two-pointer method to compare characters.
- Attempt to remove one character in the event of a mismatch.
- Check both possibilities (removing the left or right character).
- Utilize a helper function to check for palindromes.

---

### Next Steps:

- **Implement the Helper Function**:
  - Ensure it correctly checks if a given substring is a palindrome.
  - It should accept the start and end indices as parameters.

- **Edge Case Testing**:
  - Write tests for edge cases to verify your implementation.

Feel confident in your approach, and proceed to implement your solution. If you encounter any issues or have further questions, feel free to ask!

---

### Type an input and press Enter.

Q- can you me peudoCode for the helper functions ?

Function (s:string):
	// check the length of the string
	n=s.length()
	// left pointer 
	left=0
	// right pointer 
	right=len(s)-1
	While left < right : 
		If s[left]!=s[right]:
			Return False 
		left-=1
		right+=1
	Return True 

Q- for the main logic now 

Functuion : validPalindrome(self, s: str) -> bool:
		// check the length of the string 
		n=len(s)
		// initiate a left pointer 
		left=0
		//inititate a righ pointer 
		right=n-1
		// traverse the s while left < right 
		While left < right : 
			If s[left]!=right:
				del s[left] 
			left+=1
			right-=1
		Return check_palindrome(s) 

Q- this is my first iteration of the solution now use applied mathematique cs proof technique to check the correctness or noncorrectness of my algorithm before the implement the actual code oky