EX_03/level-2/rip/README.md

# The Parentheses Balancing Problem (RIP Exercise)

## What is this exercise about?

Imagine you have a string of parentheses like this: `"(()"`. 

Think of parentheses like **doors** in a building:
- `(` is like **opening a door**
- `)` is like **closing a door**

For everything to work properly, every door you open must have a matching door that closes it!

## The Problem

Sometimes we get strings where the doors don't match properly. For example:
- `"((("` - We opened 3 doors but never closed any!
- `"())"` - We opened 1 door, closed 1 door, but then tried to close another door that was never opened!

Our job is to **fix** these strings by removing the minimum number of parentheses (but replacing them with spaces instead of deleting them completely).

## Real-World Example

Let's say you have: `"(()"`

This means:
1. Open door 1: `(`
2. Open door 2: `(`  
3. Close door 2: `)`
4. But door 1 is still open! 😱

To fix this, we need to either:
- Remove one of the opening doors: ` ()` (remove first `(`)
- Or remove one of the opening doors: `( )` (remove second `(`)

Both solutions work!

## How Our Program Works

### Step 1: Count the Problems

```c
// This function counts how many parentheses we need to remove
void calculate_removals(char *str, int *left_rem, int *right_rem)
{
    int left = 0;   // Count of unmatched opening parentheses
    int right = 0;  // Count of unmatched closing parentheses
    
    // Go through each character
    while (str[i])
    {
        if (str[i] == '(')
            left++;      // Found an opening door
        else if (str[i] == ')')
        {
            if (left > 0)
                left--;  // Found a closing door for an open one
            else
                right++; // Found a closing door with no matching open door
        }
    }
}
```

**Example with `"(()"`:**
- Start: `left = 0`, `right = 0`
- See `(`: `left = 1` (one unmatched opening)
- See `(`: `left = 2` (two unmatched openings)  
- See `)`: `left = 1` (one opening got matched)
- End: `left = 1`, `right = 0`

So we need to remove **1 opening parenthesis**.

### Step 2: Try All Possible Solutions

Think of it like trying different combinations:

```c
// This function tries removing different parentheses to find all valid solutions
void solve(char *s, int pos, int left_rem, int right_rem, int open, char *path)
{
    // If we've looked at all characters
    if (pos == len)
    {
        // Check if we removed exactly what we needed and everything balances
        if (left_rem == 0 && right_rem == 0 && open == 0)
            add_result(results, path);  // This is a valid solution!
        return;
    }
    
    // Try removing this parenthesis (replace with space)
    if (s[pos] == '(' && left_rem > 0)
    {
        path[pos] = ' ';  // Replace with space
        solve(s, pos + 1, left_rem - 1, right_rem, open, path);
    }
    
    // Try keeping this parenthesis
    path[pos] = s[pos];
    if (s[pos] == '(')
        solve(s, pos + 1, left_rem, right_rem, open + 1, path);
    // ... and so on
}
```

### Step 3: Avoid Duplicates

We use a list to store all unique solutions:

```c
// This makes sure we don't print the same solution twice
void add_result(t_result **results, char *str)
{
    // Check if we already have this solution
    current = *results;
    while (current)
    {
        if (ft_strcmp(current->str, str) == 0)
            return;  // Already have it, don't add again
        current = current->next;
    }
    
    // It's new, so add it to our list
    new->str = ft_strdup(str);
    new->next = *results;
    *results = new;
}
```

## Example Walkthrough

Let's trace through `"(()"` step by step:

### Initial Analysis:
- Need to remove 1 opening parenthesis (`left_rem = 1`)
- Need to remove 0 closing parentheses (`right_rem = 0`)

### Trying Different Positions:

**Position 0 (first `(`):**
- Try removing it: `" ()"` 
  - Check: removed 1 opening ✓, no unmatched doors ✓
  - **Valid solution!** ✅

**Position 1 (second `(`):**  
- Try removing it: `"( )"` 
  - Check: removed 1 opening ✓, no unmatched doors ✓
  - **Valid solution!** ✅

**Position 2 (the `)`):**
- Can't remove it (we need to remove openings, not closings)

### Final Output:
```
 ()
( )
```

## More Complex Example: `"()())()"`

This string has:
- 3 opening doors: `(`, `(`, `(`
- 4 closing doors: `)`, `)`, `)`, `)`

So we have 1 extra closing door that needs to be removed.

**All possible solutions:**
- Remove closing at position 2: `"()( )()"`
- Remove closing at position 3: `"()() ()"`  
- Remove closing at position 4: `"( ())()"`

## Why This Exercise is Useful

This problem teaches us:
1. **Problem-solving**: Break down complex problems into smaller steps
2. **Recursion**: Try all possibilities systematically
3. **Data structures**: Use lists to store and manage results
4. **Optimization**: Find the minimum changes needed

It's like being a **door inspector** - you need to make sure every building (string) has properly matched doors (parentheses) with the minimum number of changes!