Bulk solve LeetCode problems (batch starting at index 0)

Author: jeremymanning

problems/11/gpt5-mini.md

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+# [Problem 11: Container With Most Water](https://leetcode.com/problems/container-with-most-water/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+I can’t share internal stream-of-consciousness or step-by-step private reasoning. I can, however, provide a concise, non-sensitive summary of the initial approach:
+
+Summary: The optimal solution uses a two-pointer technique. Place one pointer at the left end and one at the right end. The area is determined by the shorter of the two heights times the width between pointers. To try to increase area, move the pointer at the shorter height inward (because moving the taller pointer cannot increase the limiting height). Repeat until pointers meet, tracking the maximum area seen.
+
+## Refining the problem, round 2 thoughts
+Refinement and important notes:
+- A brute-force approach checks all pairs (O(n^2)) but is too slow for n up to 10^5.
+- Two-pointer method is O(n) time and O(1) space: start left = 0, right = n-1; at each step compute area = min(height[left], height[right]) * (right - left). Update max. Move the pointer at the smaller height inward (left++ if height[left] < height[right], else right--). If equal heights, moving either is fine; typically move one (e.g., right--).
+- Edge cases: heights of zero, all equal heights, very large n — method handles these naturally.
+- Complexity: O(n) time, O(1) extra space.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        """
+        Two-pointer approach:
+        - Initialize left pointer at 0 and right pointer at len(height) - 1.
+        - Compute area = min(height[left], height[right]) * (right - left).
+        - Update max_area.
+        - Move the pointer at the smaller height inward (since the limiting height
+          is the smaller one; moving the taller won't help increase area).
+        - Repeat until left >= right.
+        Time: O(n), Space: O(1).
+        """
+        left, right = 0, len(height) - 1
+        max_area = 0
+
+        while left < right:
+            h_left = height[left]
+            h_right = height[right]
+            width = right - left
+            area = min(h_left, h_right) * width
+            if area > max_area:
+                max_area = area
+
+            # Move the pointer at the shorter line inward
+            if h_left < h_right:
+                left += 1
+            else:
+                right -= 1
+
+        return max_area
+```
+- Notes:
+  - Approach: Two-pointer greedy-style. By always moving the shorter pointer, we potentially find a taller boundary that can increase area despite the reduced width; moving the taller pointer cannot increase the limiting height.
+  - Time complexity: O(n), as each pointer moves inward at most n steps total.
+  - Space complexity: O(1) additional space (in-place, only a few variables used).

problems/36/gpt5-mini.md

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+# [Problem 36: Valid Sudoku](https://leetcode.com/problems/valid-sudoku/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+I need to verify three constraints (rows, columns, 3x3 sub-boxes) for a 9x9 board. The board is partially filled and uses '.' for empties, so only filled cells matter. The straightforward idea: iterate every cell and for each non '.' check whether the digit has already appeared in the same row, column, or box. Using sets to track seen digits is natural. For boxes I can compute an index like (r//3, c//3) or flatten to single index 3*(r//3) + (c//3). If any duplicate is found return False early. This is simple and fast.
+
+## Refining the problem, round 2 thoughts
+Refinements and edge cases:
+- The constraints guarantee 9x9 and only digits '1'..'9' or '.', so I don't need extra validation, but code can be robust if desired.
+- Use three arrays of sets: rows[9], cols[9], boxes[9]. When visiting cell (r,c) with value v:
+  - if v in rows[r] return False
+  - if v in cols[c] return False
+  - box index b = (r//3)*3 + (c//3); if v in boxes[b] return False
+  - otherwise add v to each set.
+- Alternative solutions: bitmasks instead of sets to reduce overhead, or do three separate scans (one for rows, one for cols, one for boxes). But a single pass with 3 sets per cell is concise and O(1) time for fixed board size.
+- Time complexity: O(81) = O(1) for fixed-size, or in general O(n^2) for an n x n board. Space complexity: O(27) sets storing up to 81 elements total → O(n^2) worst-case; for fixed board it's constant.
+- Implementation details: compute box index carefully; early exit on violation.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def isValidSudoku(self, board: List[List[str]]) -> bool:
+        # rows[i] contains digits seen in row i
+        rows = [set() for _ in range(9)]
+        # cols[j] contains digits seen in column j
+        cols = [set() for _ in range(9)]
+        # boxes[k] contains digits seen in 3x3 box k, where k = (i//3)*3 + (j//3)
+        boxes = [set() for _ in range(9)]
+
+        for i in range(9):
+            for j in range(9):
+                val = board[i][j]
+                if val == '.':
+                    continue
+
+                # Check row
+                if val in rows[i]:
+                    return False
+                rows[i].add(val)
+
+                # Check column
+                if val in cols[j]:
+                    return False
+                cols[j].add(val)
+
+                # Check 3x3 box
+                box_index = (i // 3) * 3 + (j // 3)
+                if val in boxes[box_index]:
+                    return False
+                boxes[box_index].add(val)
+
+        return True
+```
+- Notes:
+  - Approach: Single pass through the board while maintaining sets for rows, columns, and 3x3 boxes. For each filled cell, check for duplicates in the corresponding row/column/box and return False on the first violation.
+  - Time complexity: O(1) for a fixed 9x9 board (practically O(81)), or O(n^2) for an n x n board. Each cell results in O(1) set operations (avg).
+  - Space complexity: O(1) for fixed board (a constant number of small sets), or O(n^2) in the worst case as the sets can store up to all elements of the board.
+  - This implementation is simple, clear, and efficient for the problem constraints.

problems/37/gpt5-mini.md

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+# [Problem 37: Sudoku Solver](https://leetcode.com/problems/sudoku-solver/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+This is a classic backtracking problem: fill each '.' with digits '1'..'9' while respecting row/column/3x3-box constraints. A naive backtracking that tries digits 1..9 for each empty cell will work but can be slow for harder puzzles. To speed up, maintain which digits are already used in each row/column/box so candidate checks are O(1). Additionally, use heuristics like choosing the empty cell with the fewest candidates (minimum remaining values, MRV) to prune the search faster. Bitmasks are a compact, fast way to represent used digits for rows/cols/boxes and to compute available candidates quickly.
+
+## Refining the problem, round 2 thoughts
+Refine to an implementation that:
+- Precomputes used digits for each row, column, and box as 9-bit masks (bits 0..8 map to digits 1..9).
+- Builds a list of empty cell coordinates.
+- Uses recursive DFS with MRV: at each step, scan the empties to pick the cell with the smallest number of candidate digits; if some cell has 0 candidates, backtrack immediately.
+- For candidate iteration, extract low bit repeatedly (mask & -mask) to iterate candidates efficiently.
+- Modify the input board in place (LeetCode requirement) and stop once a solution is found.
+
+Complexity: worst-case exponential in number of empties (m), roughly O(9^m), but MRV + constraint propagation prunes heavily. Space complexity: O(m) for recursion and empties list plus O(1) for bitmasks.
+
+## Attempted solution(s)
+```python
+class Solution:
+    def solveSudoku(self, board: list[list[str]]) -> None:
+        """
+        Modify board in-place to fill the Sudoku solution.
+        """
+        # Helper: get box index
+        def box_index(r: int, c: int) -> int:
+            return (r // 3) * 3 + (c // 3)
+
+        # Bit representation: bit 0 -> digit '1', bit 1 -> '2', ..., bit 8 -> '9'
+        rows = [0] * 9
+        cols = [0] * 9
+        boxes = [0] * 9
+        empties = []
+
+        # Initialize masks and empties list
+        for r in range(9):
+            for c in range(9):
+                ch = board[r][c]
+                if ch == '.':
+                    empties.append((r, c))
+                else:
+                    d = int(ch) - 1
+                    bit = 1 << d
+                    rows[r] |= bit
+                    cols[c] |= bit
+                    boxes[box_index(r, c)] |= bit
+
+        FULL_MASK = (1 << 9) - 1  # 0x1FF, bits for digits 1..9
+
+        # DFS with MRV (choose empty cell with minimal candidates)
+        def dfs() -> bool:
+            # If no empties left, solved
+            if not empties:
+                return True
+
+            # Find the empty cell with the fewest candidates
+            best_i = -1
+            best_count = 10
+            best_mask = 0
+            for i, (r, c) in enumerate(empties):
+                if board[r][c] != '.':
+                    continue  # already filled by earlier step
+                mask = ~(rows[r] | cols[c] | boxes[box_index(r, c)]) & FULL_MASK
+                cnt = mask.bit_count()
+                if cnt == 0:
+                    return False  # dead end
+                if cnt < best_count:
+                    best_count = cnt
+                    best_i = i
+                    best_mask = mask
+                    if cnt == 1:
+                        break  # can't get better than 1
+
+            # best_i should be valid
+            r, c = empties[best_i]
+            bidx = box_index(r, c)
+
+            # Try each candidate digit (iterate bits)
+            mask = best_mask
+            while mask:
+                bit = mask & -mask  # lowest set bit
+                d = (bit.bit_length() - 1)  # 0-based digit index
+                # place digit
+                board[r][c] = str(d + 1)
+                rows[r] |= bit
+                cols[c] |= bit
+                boxes[bidx] |= bit
+
+                # recurse
+                if dfs():
+                    return True
+
+                # backtrack
+                board[r][c] = '.'
+                rows[r] &= ~bit
+                cols[c] &= ~bit
+                boxes[bidx] &= ~bit
+
+                mask &= mask - 1  # clear lowest set bit
+
+            return False
+
+        # Kick off DFS
+        dfs()
+```
+- Notes:
+  - Approach: Depth-first search (backtracking) with bitmasking for quick validity checks and MRV (choose the empty with fewest candidates) for pruning.
+  - Time complexity: worst-case exponential, O(9^m) where m is the number of empty cells. Practical performance is far better due to constraint propagation and MRV.
+  - Space complexity: O(m) for the recursion stack and empties list, plus O(1) for the 27 masks (rows/cols/boxes).
+  - Implementation details:
+    - Use integer bitmasks for rows/cols/boxes to make candidate computations and updates O(1).
+    - Use mask & -mask and bit_length() to iterate over possible digits quickly.
+    - Modify the input board in-place as required. The dfs stops when a valid filling is found.

problems/38/gpt5-mini.md

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+# [Problem 38: Count and Say](https://leetcode.com/problems/count-and-say/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+I know this problem defines the sequence where each term describes the previous term by run-length encoding (RLE). The base is "1". For small n we can generate iteratively: start from "1" and repeatedly describe the current string to produce the next. A recursive definition is possible (countAndSay(n) = describe(countAndSay(n-1))) but recursion isn't necessary and iterative is straightforward and safer for stack/overhead.
+
+Key operations: scanning a string and counting consecutive identical digits, then appending "<count><digit>" to form the next string. Strings in Python are immutable, so building with repeated concatenation would be O(k^2); better to accumulate pieces in a list and join at the end.
+
+Constraints: n up to 30 — the strings grow but remain reasonable. So iterative approach with run-length scanning is perfect.
+
+## Refining the problem, round 2 thoughts
+Edge cases: n = 1 should return "1". For general n, loop n-1 times. Implementation details:
+- Use an index-based scan to find runs of identical characters.
+- Append count (as string) then the digit to a list, join to produce the next string.
+- Time complexity: we generate every intermediate term up to n; cost is proportional to total characters processed. Each step processes the previous string once. The length grows but for n ≤ 30 it's fine. Using lists prevents quadratic string-building cost.
+Alternative: could use itertools.groupby to simplify grouping, but manual scan is clear and explicit.
+
+## Attempted solution(s)
+```python
+class Solution:
+    def countAndSay(self, n: int) -> str:
+        if n == 1:
+            return "1"
+        
+        s = "1"
+        # build up to nth term iteratively
+        for _ in range(1, n):
+            res_chars = []
+            i = 0
+            while i < len(s):
+                j = i + 1
+                # count run length of s[i]
+                while j < len(s) and s[j] == s[i]:
+                    j += 1
+                count = j - i
+                # append count then the digit
+                res_chars.append(str(count))
+                res_chars.append(s[i])
+                i = j
+            s = "".join(res_chars)
+        return s
+```
+- Approach: Iteratively generate the sequence from 1 up to n by run-length encoding each term to get the next.
+- Time complexity: O(sum_{k=1..n} L_k) where L_k is the length of the k-th term. Practically, each step scans the previous string once, so per-step cost is O(L_{k-1}). For constraints (n ≤ 30) this is efficient. If we denote L_n as the length of nth string, the runtime is O(n * L_n) in a simplified bound.
+- Space complexity: O(L_n) extra space for building the next string (res_chars / joined string). The algorithm uses O(1) additional scalar space besides the strings being constructed.

problems/40/gpt5-mini.md

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+# [Problem 40: Combination Sum II](https://leetcode.com/problems/combination-sum-ii/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+I see this is a classic backtracking / combination generation problem where each candidate can be used at most once and we must avoid duplicate combinations. The immediate idea: sort the candidates so equal numbers are adjacent; then do a DFS/backtracking that picks or skips each candidate, but when iterating in one recursion level skip duplicates (i > start and candidates[i] == candidates[i-1]) to avoid generating identical combinations. Also, sorting allows pruning: once a candidate exceeds the remaining target, break the loop. Because target is small (<= 30) and candidates are small (<= 50), pruning will be effective. Need to ensure we pass i+1 when recursing since each number can be used only once.
+
+## Refining the problem, round 2 thoughts
+Refinements and edge cases:
+- If candidates list is empty (not in constraints, but handle gracefully) -> return [].
+- If target is 0 (not in constraints either) -> return [[]] if allowed; for this problem target >=1 so not a concern.
+- Duplicates in candidates are the main complication; sorting + skipping duplicates at the same recursion level (i > start and candidates[i] == candidates[i-1]) handles that.
+- Complexity: worst-case exponential because we explore subsets. Sorting costs O(n log n). Backtracking generates combinations with pruning; worst-case ~O(2^n) time, but practical cost is much lower given target bounds.
+- Alternative approaches: using counting of values (value->count) and choosing how many of each to include; that can also avoid duplicates but is more work and unnecessary here.
+
+I'll implement a standard sorted backtracking solution with early pruning and duplicate skipping.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
+        candidates.sort()
+        res = []
+        path = []
+        n = len(candidates)
+        
+        def backtrack(start: int, remain: int):
+            if remain == 0:
+                res.append(path.copy())
+                return
+            # iterate candidates starting from 'start'
+            for i in range(start, n):
+                # skip duplicates at the same recursion level
+                if i > start and candidates[i] == candidates[i - 1]:
+                    continue
+                val = candidates[i]
+                # pruning: if current value exceeds remaining target, no need to proceed further
+                if val > remain:
+                    break
+                # choose candidates[i]
+                path.append(val)
+                backtrack(i + 1, remain - val)  # i+1 because each number can be used only once
+                path.pop()
+        
+        backtrack(0, target)
+        return res
+```
+- Notes about the solution approach:
+  - We sort the input to make it easy to skip duplicates and to prune the search early.
+  - The recursion explores combinations by choosing each candidate at most once (hence backtrack(i+1, ...)).
+  - The check "if i > start and candidates[i] == candidates[i-1]" ensures we don't start a branch with the same number multiple times at the same recursion depth, which removes duplicate combinations.
+  - Complexity:
+    - Time: O(n log n) for sorting plus the backtracking exploration. The backtracking is exponential in the worst case — roughly O(2^n) subsets — but pruning (value > remaining target) and the small target bound often reduce actual work significantly.
+    - Space: O(n) recursion depth (plus O(k) for each combination stored in the worst case), so O(n) auxiliary space aside from the output.