bracket-sequence-restoration

{"bytes_written": 4724, "dirs_created": true}

I'll solve this step by step.

First, let me understand the problem:
- We have a string with parentheses and question marks
- We need to determine if there's exactly ONE way to replace '?' with '(' or ')' to form a valid RBS
- If unique → "YES", otherwise → "NO"

Let me think about the algorithm:

1. For a string to be a valid RBS:
   - Total count of '(' must equal total count of ')'
   - At any prefix, count of '(' ≥ count of ')'

2. For uniqueness:
   - If there's only one valid way to replace all '?'s, then YES
   - If there are multiple valid ways, then NO

Key insight: We need to check if there's exactly one valid assignment of '?' characters.

Let me implement this solution:

→ write_file({"path": "/workdir/solution.py", "content": "def solve():\n    import sys\n    input = sys.stdin.read().split()\n    t = int(input[0])\n    index = 1\n    results = []\n    \n    for _ in range(t):\n        s = input[index]\n        index += 1\n        \n        # Count known brackets and question marks\n        n = len(s)\n        open_count = s.count('(')\n        close_count = s.count(')')\n   …)

{"bytes_written": 17, "dirs_created": true}

Let me test this with the sample input:

→ write_file({"path": "/workdir/test_input.txt", "content": "3\n()()\n(??)\n((?)\n"})

{"output": "YES\nNO\nYES", "exit_code": 0, "error": null}

{"bytes_written": 4741, "dirs_created": true}

The sample output matches. But let me think more carefully about edge cases. The logic "if q_count >= 2 and both need_open > 0 and need_close > 0 then NO" might not always be correct - we need to ensure there are actually multiple valid sequences.

Let me reconsider: if we have at least 2 question marks and need both types, then we can swap the assignments of any two '?' that need different brackets, and both sequences would be valid (since the balance at each position would still be non-negative, just with different bracket types).

But wait - this isn't always true. Consider `(?` - this has 1 '?', need_open=0, need_close=1, so it would be "YES" (correct). But what about `?(` - this has 1 '?', need_open=1, need_close=0, so "YES" (correct).

Let me think about a case like `??` with length 2: need_open=1, need_close=1, q_count=2. Our solution says NO, which is correct since both `()` and `)(` - but wait, `)(` is not valid! So actually only `()` works, so it should be YES?

Let me reconsider: for `??`:
- We need 1 open and 1 close
- Possible assignments: `()`, `)(`
- `()` is valid RBS
- `)(` is NOT valid RBS (starts with close)
- So only 1 valid way → YES

But my current solution would say NO for `??` because q_count=2 and both need_open>0 and need_close>0.

Let me fix this:

→ write_file({"content": "def solve():\n    import sys\n    input = sys.stdin.read().split()\n    t = int(input[0])\n    index = 1\n    results = []\n    \n    for _ in range(t):\n        s = input[index]\n        index += 1\n        \n        n = len(s)\n        \n        # For a valid RBS, length must be even\n        if n % 2 != 0:\n            results.append(\"NO\")\n            continue\n            \n   …)

{"bytes_written": 7735, "dirs_created": true}

Wait, I need to reconsider. Let me think about `??` again:
- We need 1 '(' and 1 ')'
- Assignments: `()` and `)(`
- `()` is valid, `)(` is not valid
- So only 1 valid assignment → should be YES

But my reasoning about "swapping two '?' of different types gives another valid assignment" is wrong. Let me think more carefully.

Actually, the issue is that swapping changes the balance at intermediate positions. For `??`:
- `()` gives balance sequence: [1, 0] - valid
- `)(` gives balance sequence: [-1, 0] - invalid (negative balance)

So the swapping argument doesn't always work.

Let me think about when there are multiple valid assignments:
- We need at least 2 '?' positions
- Both types are needed (need_open > 0 and need_close > 0)
- There exist at least two different valid assignments

For `??`, there's only one valid assignment, so it should be YES.

Let me think about `(?`:
- Length 2, need_open=0, need_close=1, q_count=1 → YES (only one way)

Let me think about `(?)`:
- Length 3, invalid (odd length) → NO

Let me think about `(??)`:
- Length 4, need_open=1, need_close=1, q_count=2
- Assignments: `()()`, `)(`)`, `()()`, `)(`)`
- `()()` → balance [1,0,1,0] - valid
- `)(`)` → balance [-1,...] - invalid
- `()()` → balance [1,2,1,0] - valid
- `)(`)` → balance [-1,...] - invalid
- Wait, let me recalculate:
  - `(??)` with positions 0,1,2,3
  - `(()))` - no, that's wrong
  - `(??)` means positions: 0='(', 1='?', 2='?', 3=')'
  - Assignments: `(()))`, `()())`, `(()))`, `()())`
  - Actually: `(( ))` → `(())` - balance [1,2,1,0] - valid
  - `()())` → `()()` - balance [1,0,1,0] - valid
  - So there are 2 valid assignments → NO

So for `(??)`, the answer should be NO.

Let me reconsider the algorithm:
1. If q_count == 0: check if valid RBS → YES if valid, NO otherwise
2. If q_count == 1: only one way to assign → YES
3. If q_count >= 2:
   - If need_open == 0 or need_close == 0: all '?' must be same type → only one way → YES
   - If need_open > 0 and need_close > 0: need to check if there are multiple valid assignments

For case 3, we need to determine if there are multiple valid assignments. The key insight is:
- If there exists at least one valid assignment (guaranteed)
- And there are at least 2 '?' positions
- And both types are needed
- Then we need to check if there's more than one valid assignment

A key observation: if we can find two different valid assignments, then the answer is NO.

Let me think about when there's only one valid assignment:
- This happens when the constraints force each '?' to a specific value
- For example, `??` where only one of the two assignments works

Actually, let me reconsider `??`:
- `??` → positions 0 and 1
- Need 1 '(' and 1 ')'
- Assignments: `()` (balance: 1,0) - valid, `)(` (balance: -1,0) - invalid
- So only 1 valid assignment → YES

But for `(??)`:
- Need 1 '(' and 1 ')' for the two '?' positions
- Assignments: `(( ))` → `(())` (balance: 1,2,1,0) - valid, `()())` → `()()` (balance: 1,0,1,0) - valid
- So 2 valid assignments → NO

The difference is that in `??`, the first position MUST be '(' to avoid negative balance, so there's only one valid assignment.
But in `(??)`, both positions can be '(' or ')' in different combinations and still be valid.

So the key is: if there's a '?' that MUST be '(' (to avoid negative balance) and another that MUST be ')', then the assignment is forced and unique.

Let me implement a more careful algorithm:

→ write_file({"content": "def solve():\n    import sys\n    input = sys.stdin.read().split()\n    t = int(input[0])\n    index = 1\n    results = []\n    \n    for _ in range(t):\n        s = input[index]\n        index += 1\n        \n        n = len(s)\n        \n        # For a valid RBS, length must be even\n        if n % 2 != 0:\n            results.append(\"NO\")\n            continue\n            \n   …)

{"output": "YES\nYES\nYES", "exit_code": 0, "error": null}

Let me test this:

→ terminal({"command": "cd /workdir && python solution.py < test_input.txt"})