Q- esm the problem honi howa “Split a String in Balanced Strings “ 

Bahii 3ana definition honii enou “ Balanced” strings howa stings 3andho nefes el 3ada mt3 ‘L’ 	and ‘R’ oky 

Bahii donc enti bsh y3tok a balanced string “s” , ame lzemin split it in some number of sibstrings such that 
—> kol substring bsh yukon “balanced” oky : 

—> el objective mt3ik enik “maximum” el number mt3 el balanced strings eli tnjm tt7sel 3lih oky 


Q- bahii 5alina nw47o 7aje honii 9abla ? 
Q- wa9teh n9olo enou string “S” balanced wle ? 
—> bahii njmo n9olo enou a string “s” howa bedo balanced can 3mlt count lkol “R” w “L” w 9ethom el zouz y9abloo b3d4hom okii : 

Q- bahi s7i7 enou el input string howa bedo balanced honi ? 
→ ey 7seb el defintion honi el input string “S” howa bedo balanced oky 

Q- oky chnowa el definition mt3 substring oky 
—> howa string : yukon mwjoud fe el string el kbir 
—> a zeda ykon el elements mt3 adjante oky 

Kema s=”Amir” eli substring ynjm yukon “Am” wle “mi” oky 

**** 

Bahi twa n7wlo narj3o lel example mt3na honii oky 
Q- what is the input and output format ? 

Input : s:string 
Output : x : int 

Q- tnjm tmchi el example w t7wel ttswer kefh yt7all honii 


Bahi el input honi howa s=”RLRRLLRLRL”

Bahii ken n5mo greedy bsh “we maximise” the split score : n7wloo dema n9so sous form 
Substring of length 2 where the first “char” bsh yukon “R” o la5er “L” w l3kezs bel lkess 
Bahi 3ana constrain enou el the two element need to be adjance oky 
S3ett mch el cas foe example from index (2(0-index) ) let7are  (6()
3ana “RRLL” donc bsh y7seb honii kol as balanced substring oky 
Ame el ba9ii bsh ykono “RL” oky : 

Donc first substring 
S[0:1] = “RL” 
Second one 
s[2:6]=”RRLL” 
Third one 
s[7:8] =”RL”
Fourth one 
s[9:10]=”RL” 

Donc a7sen can enik split it into 4 substring oky 

Q- trh example a5er oky 



Fe example h4a 3ana : s=”RLRRRLLRLL” oky 

Dema n7eb net3ml analogy when i work with example ? h4a 3leh el logic bsh yukon kema hka 

Netswaro bsh yukon 3ana a linear line (table ) where fema zouz anw3 mt3 mkla oki … 
3ana 7aje esh “pizza” la5er esmha “burger” okii : 
Donc 3ana kel monter bsh yt7rk men esar lel emine oky 

Fe kol mara bsh y7wel y3ml el operation ha4i oky : 
→ enou kol mykel (point to ) nefes el 3de men pizza o burger 
(unjm ykel 1 pizza o direct be3ed 1 burger ) 
(unjm ykel 3 pizza o direct be3ed 1 burger w b3ed 1 pizza  o b3ed 3 bugger ) 
—> el 7asilo yo9eff ken fek ysol nefess the 3dede men pizza o burger oky 
—> kef yodel awl 7aje bsh : 
Y3ml update lel “counter” oky 
W b3ed direct bsh yrj3 ykel (parcour men a5er plase +1) oky 

Q- donc the logic honi  bsh yukon “dynamique travzerse oky “ 
Donc ntswer pointer i and j y3wnoo honii oky bsh dema n3rf el bound el scanerha 
Q- also fe kol etape el monster bsh yt3da beha lzemna n7sebo 9deh men pizza wle burger 3ana klemna okii 


5lina nwlo ntswrohoa fe example 

3ana awl 7aje counter =0 // farg men loul oky 



El table mt3na bsh tkon s=”RLRRRLLRLL” oky 
 Bshi el monster bsh ykel “R” → s[0] 
Honi kle ken “R” donc mzelna mnmoch n9foo oky 
Bash ykel twa “L” → s[1] 
Honi kle “L” twa donc mswolna le awl mara el monster eat nfes el 3ded men “R” and “L” oky 
Donc njmo n3mlo update lel  counter oky bsh ywlii +1 
Twa el monster bsh ywlii ybda ykle men s[1+1] oky 
Honi howa kel ken “R” donc mzel mnjmoch n9foo oky 
Bash ykel twa zeda “R” donc mzel mnjmoch n9foo oky 
Bsh ykel twa zeda “R” donc mzel mnjmoch n9foo oky 
Twa bsh ykel awl mara “L” oky 
Twa bsh ykel zeda mara o5ra “L” oky 
(just bsh mnswch  tawa 3ana enou howa kle 3 marat “R” o martin “L” oky ) 

Twa bsh yklel “R” donc mnjmoch nafo twa 
Twa bsh ykel “L” kef kef njmo nzedo oky 
Twa bsh ykel “L” donc 
→ honi el monster kel awl mara ykel nfess el 3ded men “R” o “L” oky 

Donc njmo n3mlo update lel counter +1 

—> donc honii bsh yukon el return counter = 2 okiii 


Q- ch9wlik twa t7well el tounsie mt3i in english 
Q- without removing my unique style here oky 

Alright! Let's translate your unique Tunisian explanation into English while preserving your style and flow.

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**Example:** We have the string `s = "RLRRRLLRLL"`. 

I like to use analogies when working with examples, so here’s how I’m thinking of it:

Let’s imagine we have a linear table (a line) where there are two types of food. We’ve got one thing called "pizza," and the other is called "burger."

Each time the monster (let’s say that’s us, moving along the line) eats, he moves from left to right. 

Here’s the rule for the monster’s eating:

→ Every time the monster (points to) eats, he eats the same number of pizzas or burgers. 
(He can eat 1 pizza, then immediately after 1 burger.)
(Or he can eat 3 pizzas, then immediately after 1 burger, then another pizza, then 3 burgers in a row.)

—> The catch is, he only stops when he’s eaten the same number of pizzas and burgers. 
—> So, when he stops the first time, we need to update our "counter."
After that, the monster will keep eating (starting again from the next place).

**Q:** So the logic here is a **dynamic traverse**.
Let's imagine pointers `i` and `j` that help us know the bounds of our scan.
Also, at every step the monster takes, we need to count how many pizzas and burgers he's eaten.

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Let’s jump into an example:

1. At the start, `counter = 0` (we begin with an empty counter).

2. Our table is `s = "RLRRRLLRLL"`.
- The monster first eats “R” → `s[0]`.
- He just ate an "R," so we can’t balance things yet (no "L"s eaten yet).
- Now he eats an “L” → `s[1]`. 
- Here, the monster has eaten an equal number of "R"s and "L"s for the first time, so we update the counter by 1.
- Now, the monster will start eating from the next position `s[1+1]`.

3. Next, the monster eats an "R" → `s[2]`.
- He ate an "R," but no balance yet.
- He eats another "R" → `s[3]`, still no balance.
- And again, another "R" → `s[4]`.

4. Now, the monster eats “L” → `s[5]`.
- Followed by another “L” → `s[6]`.
- (Don’t forget, by this point the monster has eaten 3 "R"s and 2 "L"s.)

5. Next, he eats another "R" → `s[7]`. 
- Still no balance.
- Then, an "L" → `s[8]`. 
- And again, an "L" → `s[9]`.

At this point, the monster has balanced his eating for the second time, so we can update the counter by +1 again.

**Final return: `counter = 2`.**

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Q- oky this this my thinking process so far 
Q- it’s first begin with tounis which my native language then translate it into english 
Q- i do not want the final solution now 
Q- just give me some feabacck about thinking proess so far 
Q- did i got the right approach 
And did i am in the right direction oky yes or non ?