459. Repeated Substring Pattern
Difficulty: Easy
Topics: String
Companies:
Problem Statement:
Given a string s, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together.
Examples:
Example 1:
     Input: s = "abab"
Output: true
Explanation: It is the substring "ab" twice.
   
content_copy Use code with caution.
Example 2:
     Input: s = "aba"
Output: false
   
content_copy Use code with caution.
Example 3:
     Input: s = "abcabcabcabc"
Output: true
Explanation: It is the substring "abc" four times or the substring "abcabc" twice.
   
content_copy Use code with caution.


Q-can you explain it using your owns words ?

Oky so we are given a string .. check if we can construct it in the same manner by taking a substring and repeated k times 

Q- can if the input/output format 
Input: string 
Output : bool 
Q- can you go through an example ? 

s=”abab” 

→ as we see a have pattern that repeat two times 
“Ab” * 2
So we can return True 

s=”aba” 
→ these now specific pattern repeat so we return False 

s=”abcabcabcabc” 
→ here we have pattern repent two times “abcabc” 
→ and “abc” 4 times times 

So we return True 

Q-first question how to extract the pattern in the first place to get it ? 

One hypothic here 
Using pattern matching when tranvsing the input string 
If we found the multiplication of the length of this substring and number of it’s occences 
Equal to the length of the input string we return true ok else false 


Q- one trick to approach this problem ? 
Is to not transverse the full string length but instead only the first halp  of the string 

Check all the potential substring in the first half 

If we found a substring that very this 
1- it length it divisible by the length of input string 
2- if the number of repeated substring is equal to the length_input_string / length_sub_string 

Example 

“Abcabc” 
→ all the poetnetional sub string 
a , ab , abc 
Check one input_string.length is disivble by sub_string_length.length 

Check two number of occurrences of this substring equal to input_string.length//sub_string_length.length 


Q- before i continue did my thinking process so far true 

Here is a well-formatted version of your text:

---

Yes, your thinking process is correct so far. You're on the right track with your approach. Let's summarize your strategy:

1. **Consider all potential substrings**:  
   Iterate over all substring lengths from 1 up to half of the input string's length. This is because a repeating pattern must be shorter than or equal to half of the string's length.

2. **Check divisibility**:  
   For each potential substring, check if the length of the input string is divisible by the length of the substring. This ensures that the substring can be repeated an integer number of times to possibly form the original string.

3. **Verify repetition**:  
   Calculate how many times the substring would need to repeat to match the length of the input string (i.e., `repeat_count = len(s) // len(substring)`). Then, create a new string by repeating the substring this number of times.

4. **Compare with the original string**:  
   If the newly formed string equals the original string, you can conclude that the original string is composed of repeated substrings, and return `True`.

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### Example:

For the string `"abcabc"`:

- Potential substrings are `"a"`, `"ab"`, and `"abc"`.
- Checking `"abc"`:
   - Length of `s` is 6, and the length of `"abc"` is 3.
   - `6 % 3 == 0`, so it divides evenly.
   - Repeat `"abc"` twice: `"abc" * 2 = "abcabc"`.
   - `"abcabc"` equals the original string, so return `True`.

---

Your two conditions are valid and form a solid basis for solving the problem:

1. **Divisibility Condition**:  
   The input string's length must be divisible by the potential substring's length.

2. **Repetition Condition**:  
   Repeating the substring `n` times (where `n = len(s) // len(substring)`) should reconstruct the original string.

---

This approach efficiently checks for all possible substrings that could form the original string through repetition. It's a practical solution that balances thoroughness with efficiency, avoiding unnecessary checks beyond half the string's length.

Feel free to proceed with implementing this logic in your code!

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Type an input and press Enter.


Q- pseudoCode : 

Function : repeatedSubstringPattern(self, s: str) -> bool:
		// get the length of the s 
		n=len(s)
		For i in range(n//2):
			sub_string=s[:i+1]
			k=len(sub_string=
			// check divibilty to found out if this substring can be repeated integer time or 
			If n%k==0:
				// now found the number of repeated 
				nb=n//k
				Return sub_string*nb==s

Q- using applied computation mathemqiue algorithmique proof to check the correctness of my solution