Q- Separate Black and White Balls 

Bahi 3andik “n” tabe ge table w7totin , for kora 3andha zozu alwenn 
—> aby4 and ak47el oky 

Bahiii yasedik as input you have a binary string “s” : 
→ 1 <- -> black ball 
→ 0 <- -> white ball 
Respectively 

Q- oky the allowed operation is each time you can choose two adjacent balls and swap them oky 

Q- your task enik trj3 el 3ded minimum mt3 steps : bsh t5ali el string grouped as 
→ all the black balls to the right 
→ all the white balls in the left 



Q- example : s =”101” 

One solution come to my mind is to iterate each time to a pair adhjecen element and check if 
I-1 ==1 and i==0 then swap them 
And update a variable that we have initiate to zero and we update it each time a make a swap 

PesudoCode : Function(s:string)->int: 
		n=s.length()
		count=0
		For i from 1 to n-1:
			If s[i-1]==1 and s[i]==0:
				s[i-1],s[i]=s[i],s[i-1] 
				count+=1
		Return count 

Q- so now i want  you to focus on my first attemption and check it correctness or incorrectness oky 
Q- i want you to see if it’s true or not using mathematical algorithm proof that check the correcntss of the problem before even implement it oky ?
				
Your initial approach to solving the problem of separating black (1) and white (0) balls by performing adjacent swaps is a good starting point. However, upon closer examination, the algorithm as currently designed does not always produce the correct minimal number of swaps required to achieve the desired arrangement. Let's delve into why this is the case and how we can adjust the approach to ensure correctness.
Understanding the Problem
Given a binary string s, where:
1 represents a black ball.
0 represents a white ball.
Objective: Arrange the string such that all 0s are on the left and all 1s are on the right, using the minimum number of adjacent swaps.
Example Analysis
Let's analyze the example you provided:
Input: s = "101"
Desired Output: "011"
Your Algorithm:
Iterate from left to right.
If you find a 1 followed by a 0, swap them and increment the count.
Execution:
Step 1: Swap the first two characters: "101" → "011", count = 1.
Result: The algorithm returns 1, which is correct in this case.
However, let's consider a more complex example to test the algorithm's correctness.
Counter-Example
Input: s = "1001"
Desired Output: "0011"
Minimal Swaps Needed: 2
Step 1: Swap indices 0 and 1: "1001" → "0101", count = 1.
Step 2: Swap indices 1 and 2: "0101" → "0011", count = 2.
Your Algorithm's Execution:
First Pass:
i = 1: s[0] = '1', s[1] = '0' → Swap → "0101", count = 1.
i = 2: s[1] = '1', s[2] = '0' → Swap → "0011", count = 2.
i = 3: s[2] = '1', s[3] = '1' → No swap.
Result: The algorithm returns 2, which is correct in this case.
However, in some other cases, the algorithm might not perform multiple necessary passes.
Potential Flaws
Consider another example:
Input: s = "1100"
Desired Output: "0011"
Minimal Swaps Needed: 4
Step-by-Step Swaps:
Swap indices 1 and 2: "1100" → "1010", count = 1.
Swap indices 2 and 3: "1010" → "1001", count = 2.
Swap indices 0 and 1: "1001" → "0101", count = 3.
Swap indices 1 and 2: "0101" → "0011", count = 4.
Your Algorithm's Execution:
First Pass:
i = 1: s[0] = '1', s[1] = '1' → No swap.
i = 2: s[1] = '1', s[2] = '0' → Swap → "1010", count = 1.
i = 3: s[2] = '1', s[3] = '0' → Swap → "1001", count = 2.
Second Pass (if implemented):
i = 1: s[0] = '1', s[1] = '0' → Swap → "0101", count = 3.
i = 2: s[1] = '1', s[2] = '0' → Swap → "0011", count = 4.
Result: If your algorithm performs only one pass, it would incorrectly return 2 instead of the required 4 swaps.
Conclusion: Your current pseudocode performs only a single pass through the string. While it works for some cases, it fails to account for scenarios where multiple passes are necessary to move all 1s past all 0s.


Q- oky my first solution solve some problem (the ones that only require one pass ) but do not work well for the case where require multiple pass ? 

Q- so can you go of a better approach here ? 

Let’s build our intuition by looking an one example ? 


 s = "1100"


→ in this example we  can see the number of swap required here 
This can be seen as the number of 1s before each time we met a zero “0” . i mean is in the number of 1s we find each time before getting “0” 


→ in this example
We have 1 + 1 = 2 (ones ) before meeting the first one oky 
Next we dones not meet any other 1s so w return 2 


Q- now can you fromlize this thinking process into an actual formal algorithm ? 


Q- what is the input/output format and the function signature ? 


In the input -> s: string 
The output -> min_swap:int 
The Function signature -> Function(s:string)->int


Q- do we need to initte some variables? 
Yes : we need bacily two variable here First one to track the ones found so far ? 
The second one to check the “min_swap” to track the minumual number of swap need it herre 


Q- how the main logic is done here 
We need to iterate throuh the string “char” per “char” 
If char ==”1”: 
		nb_ones+=1 // we have found one 
Else : 
		min_swap+=nb_ones // now we have found (if possible ) x number of ones and then a “zero” to move them to the right we need at least nb_ones steps oky 


Return min_swap


Q- without given any feedback just genelize the peudoCode here 




Algorithm:
Initialize nb_ones to 0 to track the number of 1s found so far.
Initialize min_swaps to 0 to track the minimum number of swaps needed.
Iterate through each character in the string s:
If the character is '1', increment nb_ones by 1.
If the character is '0', increment min_swaps by the current count of nb_ones.
Return min_swaps.




Q- now be my coding mentor : 
→ give your honest feedback about my thinking evolution 
How to navigate this problem given my notes 
Try to explain some pattern in my thinking process i do know 
→ next you are mathematiczl genius so try to reformalte my thinking process into more abstract notation and fromalization 
That combine the power of math and cs oky 


→ then compare the O() time & space complexity oky