Add solutions and explanations for problems 237, 797, 1008, 1079, 1261, 1347, 1551, 1605, 1630, 1823, 2120, 2221, 2326, 2482, 2785, 3562 #119
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…1, 1347, 1551, 1605, 1630, 1823, 2120, 2221, 2326, 2482, 2785, 3562
Reviewer's GuideAdds accepted solutions and detailed explanations for 16 LeetCode problems, refines several existing explanations to a more constraint‑ and approach‑oriented style, and adjusts multiple solution implementations for clarity, small optimizations, and consistency with repo conventions, including introducing a tree-DP/knapsack solution for problem 3562. Flow diagram for tree DP knapsack in maxProfit (problem 3562)flowchart TD
start_main["Start maxProfit(n, present, future, hierarchy, budget)"] --> build_adj["Build adj_list from hierarchy (0-indexed)"]
build_adj --> call_dfs_root["Call dfs(employee=0, has_discount=False)"]
call_dfs_root --> dfs_start
subgraph dfs_subgraph["dfs(employee, has_discount)"]
dfs_start["Enter dfs(employee, has_discount)"] --> compute_cost["If has_discount: cost = present[employee] / 2; else cost = present[employee]"]
compute_cost --> compute_profit["profit = future[employee] - cost"]
compute_profit --> init_buy_skip
init_buy_skip["Initialize buy_current and skip_current"] --> check_afford
check_afford{"cost <= budget?"} -->|Yes| set_buy["buy_current = { cost: profit }"]
check_afford -->|No| empty_buy["buy_current = {}"]
set_buy --> init_skip["skip_current = { 0: 0 }"]
empty_buy --> init_skip
init_skip --> children_loop_start
children_loop_start["For each child in adj_list[employee]"] --> children_loop_cond{"More children?"}
children_loop_cond -->|No| merge_buy_skip
children_loop_cond -->|Yes| process_child["Take next child"]
process_child --> dfs_child_disc["child_with_discount = dfs(child, True)"]
dfs_child_disc --> dfs_child_nodisc["child_no_discount = dfs(child, False)"]
dfs_child_nodisc --> merge_buy_child
merge_buy_child["Merge buy_current with child_with_discount using knapsack"] --> merge_buy_inner
merge_buy_inner["For each (spent, prof) in buy_current and (child_spent, child_prof) in child_with_discount:<br/>if spent + child_spent <= budget:<br/>new_buy[spent + child_spent] = max(existing, prof + child_prof)"] --> update_buy["buy_current = new_buy"]
update_buy --> merge_skip_child
merge_skip_child["Merge skip_current with child_no_discount using knapsack"] --> merge_skip_inner
merge_skip_inner["For each (spent, prof) in skip_current and (child_spent, child_prof) in child_no_discount:<br/>if spent + child_spent <= budget:<br/>new_skip[spent + child_spent] = max(existing, prof + child_prof)"] --> update_skip["skip_current = new_skip"]
update_skip --> children_loop_cond
merge_buy_skip["Merge buy_current and skip_current into result"] --> result_loop["For each (spent, prof) in buy_current and skip_current:<br/>result[spent] = max(existing, prof)"]
result_loop --> dfs_return["Return result (dict: spent_budget -> max_profit)"]
end
dfs_return --> main_collect["result = dfs(0, False)"]
main_collect --> main_answer["answer = max(result.values()) if result else 0"]
main_answer --> end_main["Return answer from maxProfit"]
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WalkthroughThis PR adds a new LeetCode problem ( Changes
Estimated code review effort🎯 4 (Complex) | ⏱️ ~60 minutes
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Hey there - I've reviewed your changes - here's some feedback:
- For problem 1008, the previous iterative stack-based reconstruction runs in linear time, while the new recursive split-and-recurse approach can degrade to O(n²) due to repeated scanning and slicing; consider restoring the stack solution or adding bounds-based recursion to keep the construction O(n).
- In the tree-DP/knapsack solution for problem 3562, the per-node dictionaries of (spent → profit) are not pruned for dominated states, so merging children is O(budget²); you can significantly reduce runtime by discarding entries where another state has both lower or equal cost and higher or equal profit.
Prompt for AI Agents
Please address the comments from this code review:
## Overall Comments
- For problem 1008, the previous iterative stack-based reconstruction runs in linear time, while the new recursive split-and-recurse approach can degrade to O(n²) due to repeated scanning and slicing; consider restoring the stack solution or adding bounds-based recursion to keep the construction O(n).
- In the tree-DP/knapsack solution for problem 3562, the per-node dictionaries of (spent → profit) are not pruned for dominated states, so merging children is O(budget²); you can significantly reduce runtime by discarding entries where another state has both lower or equal cost and higher or equal profit.
## Individual Comments
### Comment 1
<location> `solutions/1551/01.py:7-13` </location>
<code_context>
class Solution:
def numTilePossibilities(self, tiles: str) -> int:
- count = Counter(tiles)
- res = 0
+ from collections import Counter
</code_context>
<issue_to_address>
**suggestion (performance):** The loop can be simplified and reduced to `n // 2` iterations with a direct formula.
Right now you still scan from 0 upward and branch on each element before breaking. Because only indices `< n // 2` can produce values `< n`, you can iterate directly with `for i in range(n // 2):` and remove the conditional/break. Or, if you don’t need per-iteration work, replace the loop with the closed form `n * n // 4` to compute the same result.
</issue_to_address>
### Comment 2
<location> `explanations/1079/en.md:8` </location>
<code_context>
-- **Space Complexity:** O(n) recursion depth + O(1) counts.
-- **Edge Case:** Single tile → exactly 1 sequence.
+* **String Length:** The tiles string has length 1-7, consisting of uppercase English letters. This small constraint allows for backtracking solutions.
+* **Time Complexity:** We generate all possible sequences using backtracking. The number of sequences depends on the character frequencies. **Time Complexity: O(2^n)** where n is the number of unique sequences, **Space Complexity: O(n)** for recursion stack and character counts.
+* **Edge Case:** If tiles has only one character, there's exactly one sequence (the character itself).
</code_context>
<issue_to_address>
**suggestion:** Clarify the time-complexity notation and what `n` represents.
Referring to `n` as “the number of unique sequences” is misleading—`n` is usually the input size (e.g., `len(tiles)`), and the number of sequences is what scales like `2^n`. Please rephrase to define `n` as the input length (e.g., `O(2^n)` where `n` is `len(tiles)`), or use a more standard bound such as `O(2^n * n)` if you want to account for sequence length as well.
</issue_to_address>Help me be more useful! Please click 👍 or 👎 on each comment and I'll use the feedback to improve your reviews.
| - **Space Complexity:** O(n) recursion depth + O(1) counts. | ||
| - **Edge Case:** Single tile → exactly 1 sequence. | ||
| * **String Length:** The tiles string has length 1-7, consisting of uppercase English letters. This small constraint allows for backtracking solutions. | ||
| * **Time Complexity:** We generate all possible sequences using backtracking. The number of sequences depends on the character frequencies. **Time Complexity: O(2^n)** where n is the number of unique sequences, **Space Complexity: O(n)** for recursion stack and character counts. |
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suggestion: Clarify the time-complexity notation and what n represents.
Referring to n as “the number of unique sequences” is misleading—n is usually the input size (e.g., len(tiles)), and the number of sequences is what scales like 2^n. Please rephrase to define n as the input length (e.g., O(2^n) where n is len(tiles)), or use a more standard bound such as O(2^n * n) if you want to account for sequence length as well.
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Actionable comments posted: 5
Caution
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⚠️ Outside diff range comments (1)
solutions/2482/01.py (1)
2-2: MissingListimport for type annotation.The type hint uses
Listbut it's not imported. While LeetCode's environment typically handles this, adding the import improves portability.+from typing import List + class Solution: def onesMinusZeros(self, grid: List[List[int]]) -> List[List[int]]:
🧹 Nitpick comments (8)
solutions/1347/01.py (1)
3-3: Consider moving import to module level.While functional, moving the
Counterimport inside the function adds a small lookup overhead on each call. Python style guidelines typically recommend module-level imports for better readability and performance.Apply this diff to move the import to module level:
+from collections import Counter + class Solution: def minSteps(self, s: str, t: str) -> int: - from collections import Counter - count_s = Counter(s) count_t = Counter(t)solutions/1008/01.py (1)
14-20: Clean recursive implementation with minor performance consideration.The recursive approach is intuitive and correct. The array slicing (
preorder[1:i],preorder[i:]) creates copies at each recursive level, resulting in O(n²) space complexity in the worst case (skewed tree). For competitive programming this is acceptable, but for production code you might consider passing start/end indices instead.If you want to optimize space complexity to O(n), consider passing indices:
class Solution: def bstFromPreorder(self, preorder: List[int]) -> Optional[TreeNode]: + def build(start: int, end: int) -> Optional[TreeNode]: + if start >= end: + return None + + root = TreeNode(preorder[start]) + i = start + 1 + while i < end and preorder[i] < preorder[start]: + i += 1 + + root.left = build(start + 1, i) + root.right = build(i, end) + return root + + return build(0, len(preorder)) - if not preorder: - return None - - root = TreeNode(preorder[0]) - - # Find the split point where values become greater than root - i = 1 - while i < len(preorder) and preorder[i] < preorder[0]: - i += 1 - - root.left = self.bstFromPreorder(preorder[1:i]) - root.right = self.bstFromPreorder(preorder[i:]) - - return rootexplanations/2482/en.md (1)
5-5: Consider using proper markdown headings instead of bold emphasis.Lines 5, 11, 21, 26 use
**bold text**as sub-headings. Using####headings would improve document structure and accessibility.-**Constraints & Edge Cases** +#### Constraints & Edge Cases-**High-level approach** +#### High-level approach-**Brute force vs. optimized strategy** +#### Brute force vs. optimized strategy-**Decomposition** +#### DecompositionAlso applies to: 11-11, 21-21, 26-26
explanations/1079/en.md (1)
8-8: Minor: Time complexity description is circular.The statement "O(2^n) where n is the number of unique sequences" is circular since we're trying to count sequences. A clearer formulation would be that the time complexity is bounded by the total number of permutations, which for input length L with character frequencies is at most O(L! / ∏(count[c]!)), or simply "exponential in the input size but manageable for L ≤ 7."
explanations/1008/en.md (1)
36-38: Clarification: Split condition should use strict inequality for general correctness.The condition
preorder[i] >= preorder[0]works due to unique values, but conceptually for BST the right subtree contains values strictly greater than the root. Consider usingpreorder[i] > preorder[0]to be more precise, matching the BST property where left < root < right.solutions/3562/01.py (2)
2-2: Unused parametern.The parameter
nis declared but never used in the method. This is acceptable if the LeetCode problem signature requires it, but worth noting. The actual number of employees can be inferred fromlen(present).If LeetCode allows modifying the signature, consider removing
n. Otherwise, you could add a comment like# n is provided but not needed; tree structure is in hierarchy.
3-4: Consider moving imports to module level.While placing imports inside functions can improve startup time for unused code paths, PEP 8 recommends module-level imports for clarity. For LeetCode solutions this is often acceptable, but consider moving them to the top.
explanations/1347/en.md (1)
18-18: Vary phrasing to improve readability.The phrase "we need to replace" is repeated in the text. Consider rewording for stylistic variety.
For example, rephrase line 18 as:
-* **Optimized:** Count character frequencies in both strings. For each character, if s has more occurrences than t, we need to replace (count_s[char] - count_t[char]) characters in t. +* **Optimized:** Count character frequencies in both strings. For each character, if s has more occurrences than t, calculate the difference (count_s[char] - count_t[char]) as the number of characters to replace in t.
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📒 Files selected for processing (33)
data/leetcode-problems.json(1 hunks)explanations/1008/en.md(1 hunks)explanations/1079/en.md(1 hunks)explanations/1261/en.md(1 hunks)explanations/1347/en.md(1 hunks)explanations/1551/en.md(1 hunks)explanations/1605/en.md(1 hunks)explanations/1630/en.md(1 hunks)explanations/1823/en.md(1 hunks)explanations/2120/en.md(1 hunks)explanations/2221/en.md(1 hunks)explanations/2326/en.md(1 hunks)explanations/237/en.md(1 hunks)explanations/2482/en.md(1 hunks)explanations/2785/en.md(1 hunks)explanations/3562/en.md(1 hunks)explanations/797/en.md(1 hunks)solutions/1008/01.py(1 hunks)solutions/1079/01.py(1 hunks)solutions/1261/01.py(1 hunks)solutions/1347/01.py(1 hunks)solutions/1551/01.py(1 hunks)solutions/1605/01.py(0 hunks)solutions/1630/01.py(1 hunks)solutions/1823/01.py(1 hunks)solutions/2120/01.py(1 hunks)solutions/2221/01.py(0 hunks)solutions/2326/01.py(1 hunks)solutions/237/01.py(0 hunks)solutions/2482/01.py(1 hunks)solutions/2785/01.py(1 hunks)solutions/3562/01.py(1 hunks)solutions/797/01.py(0 hunks)
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🪛 LanguageTool
explanations/2482/en.md
[style] ~7-~7: As an alternative to the over-used intensifier ‘very’, consider replacing this phrase.
Context: ...st 10^5. This means the matrix could be very wide or very tall. * Time Complexity: We...
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[style] ~18-~18: You have already used this phrasing in nearby sentences. Consider replacing it to add variety to your writing.
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solutions/3562/01.py
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⏰ Context from checks skipped due to timeout of 90000ms. You can increase the timeout in your CodeRabbit configuration to a maximum of 15 minutes (900000ms). (1)
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🔇 Additional comments (29)
solutions/2785/01.py (1)
5-5: LGTM! Minor comment improvement.The updated comment is more concise while maintaining clarity.
solutions/1823/01.py (1)
3-7: LGTM! Clean Josephus problem implementation.The implementation correctly uses the iterative Josephus formula with proper 1-indexed conversion.
solutions/1347/01.py (1)
5-13: LGTM! Correct anagram step calculation.The logic correctly counts excess characters in
sthat need replacement to form an anagram oft.solutions/1261/01.py (1)
13-18: LGTM! Good use of early return pattern.The refactoring to use early return improves readability while maintaining the same logic. The tree recovery algorithm correctly assigns values using the formula
2*val+1for left and2*val+2for right children.solutions/2326/01.py (2)
14-19: LGTM! Cleaner traversal logic.The refactoring eliminates the redundant
currentvariable and directly useshead, making the code more concise while maintaining correctness.
21-32: LGTM! Correct spiral traversal implementation.The direction-change logic properly handles boundary conditions and already-filled cells for spiral matrix traversal.
solutions/2120/01.py (1)
11-23: LGTM! Valid refactoring to explicit conditionals.The change from dictionary-based direction mapping to explicit conditionals makes the movement logic more transparent. The boundary checking and counting logic remain correct.
data/leetcode-problems.json (1)
25843-25843: JSON entry properly formatted and consistent with existing structure.The new problem entry for 3562 follows the established conventions: correct field structure, appropriate difficulty level, valid LeetCode problem URL, and empty tags array consistent with other entries. The comma placement on line 25842 correctly precedes this entry.
solutions/1551/01.py (1)
1-15: LGTM!The algorithm correctly computes minimum operations using the symmetry property of the array. Since elements form an arithmetic sequence
[1, 3, 5, ..., 2n-1]with averagen, summing differences from the lower half is optimal. The early break whencurrent >= targetavoids unnecessary iterations.solutions/1079/01.py (1)
1-16: LGTM!The backtracking approach with return-value accumulation is clean and correct. Each available character creates a new sequence (+1), then recursively explores extensions. The Counter handles duplicate tiles properly.
solutions/2482/01.py (1)
15-23: LGTM!Good optimization moving
zerosRowcalculation outside the inner loop. The preallocated matrix with direct assignment is cleaner than dynamic row construction.solutions/1630/01.py (1)
5-21: LGTM!The arithmetic progression check is clean and correct. The slice on line 6 creates a copy, so the in-place sort doesn't affect the original
numsarray. Early break on mismatch is a good optimization.explanations/2482/en.md (1)
54-67: Trace table is well-structured and accurate.The cell-by-cell walkthrough clearly demonstrates the algorithm. Verified calculations match the expected output
[[0,0,4],[0,0,4],[-2,-2,2]].explanations/2326/en.md (1)
5-59: [Covered by consolidated markdown heading refactor comment above.]Content quality and structure are good. Apply the same ### heading refactor as noted in the consolidated comment for all 8 files.
explanations/1551/en.md (1)
5-50: [Covered by consolidated markdown heading refactor comment above.]Content and examples are clear. Apply the ### heading refactor.
explanations/2221/en.md (1)
5-50: [Covered by consolidated markdown heading refactor comment above.]Content is clear with good iteration trace. Apply the ### heading refactor.
explanations/1261/en.md (1)
5-52: [Covered by consolidated markdown heading refactor comment above.]Content explains recovery and querying well. Apply the ### heading refactor.
explanations/797/en.md (1)
5-62: [Covered by consolidated markdown heading refactor comment above.]Content explains DFS and backtracking clearly. Apply the ### heading refactor.
explanations/1605/en.md (1)
5-57: [Covered by consolidated markdown heading refactor comment above.]Content is clear with helpful verification section. Apply the ### heading refactor.
explanations/2120/en.md (1)
5-57: [Covered by consolidated markdown heading refactor comment above.]Content explains per-position simulation clearly. Apply the ### heading refactor to complete standardization.
explanations/2785/en.md (1)
1-56: LGTM!The explanation is well-structured with clear constraints, complexity analysis, and a detailed trace walkthrough. The example correctly demonstrates the vowel sorting algorithm with proper ASCII ordering (uppercase before lowercase).
explanations/237/en.md (1)
1-49: LGTM!The explanation clearly describes the classic "copy next node's value and skip" technique for deleting a node without head access. The trace walkthrough accurately demonstrates the algorithm with the example.
explanations/1079/en.md (1)
44-55: LGTM!The trace walkthrough correctly enumerates all 8 possible sequences for "AAB" and demonstrates the backtracking approach clearly.
explanations/1823/en.md (1)
1-53: LGTM!The Josephus recurrence formula and step-by-step calculation are mathematically correct. The trace table clearly demonstrates the computation for n=5, k=2 leading to the correct result of 3 (1-indexed).
explanations/1008/en.md (1)
1-56: LGTM overall!The recursive split-point approach is clearly explained, and the trace walkthrough correctly demonstrates building the BST from
[8,5,1,7,10,12].explanations/3562/en.md (1)
43-53: LGTM!The trace walkthrough clearly demonstrates the tree DP + knapsack approach with correct profit calculations for both buy and skip options.
solutions/3562/01.py (2)
11-12: Verifylru_cacheusage with dict returns.Using
@lru_cacheondfswhich returns adictis technically valid since caching is based on input arguments, not return values. However, ensure callers never mutate the returned dictionaries, as that would corrupt cached results. In this code, the returned dicts are only read in iterations, so this appears safe.
6-65: Algorithm logic looks correct.The tree DP + knapsack approach correctly:
- Builds the adjacency list from hierarchy
- Processes each employee with/without discount states
- Merges children results via knapsack convolution
- Returns the maximum profit across all budget allocations
explanations/1347/en.md (1)
3-48: Content and structure align well with PR standardization objectives.The explanation now follows a clear, consistent structure with explicit constraints, complexity analysis, high-level strategy, decomposition steps, and a concrete example walkthrough. The technical content is accurate—the frequency-counting approach is sound, the example correctly demonstrates the concept, and the time/space complexity analysis (O(n) time, O(1) space for 26-letter alphabet) is correct.
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| **Restate the problem** | ||
| Given two equal-length lowercase strings `s` and `t`, find the minimum number of character replacements in `t` to make it an anagram of `s`. | ||
| **Constraints & Edge Cases** |
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Use Markdown heading syntax for section headers instead of bold emphasis.
Lines 5, 11, 15, and 20 use bold text (e.g., **Constraints & Edge Cases**) for subsection headers. Per Markdown conventions and linting rules, these should use heading syntax (###) to maintain proper document structure and consistency.
Apply this diff to convert bold emphasis to proper Markdown headings:
-**Constraints & Edge Cases**
+### Constraints & Edge CasesRepeat for lines 11, 15, and 20:
**High-level approach**→### High-level approach**Brute force vs. optimized strategy**→### Brute force vs. optimized strategy**Decomposition**→### Decomposition
Also applies to: 11-11, 15-15, 20-20
🧰 Tools
🪛 markdownlint-cli2 (0.18.1)
5-5: Emphasis used instead of a heading
(MD036, no-emphasis-as-heading)
🤖 Prompt for AI Agents
In explanations/1347/en.md around lines 5, 11, 15, and 20, subsection headers
are written with bold emphasis (e.g., **Constraints & Edge Cases**) instead of
Markdown heading syntax; replace each bolded subsection header with a level-3
Markdown heading by changing the bold text to start with "### " (so line 5
becomes "### Constraints & Edge Cases", line 11 "### High-level approach", line
15 "### Brute force vs. optimized strategy", and line 20 "### Decomposition")
ensuring spacing consistency and leaving surrounding content unchanged.
| **Constraints & Edge Cases** | ||
|
|
||
| **1.1 Constraints & Complexity** | ||
| - **Input Size:** `n, m <= 500`. | ||
| - **Time Complexity:** O(m * k log k), where k is subarray length (sort each subarray). | ||
| - **Space Complexity:** O(k) per query for the sorted copy. | ||
| - **Edge Case:** Subarray of length 2 is always arithmetic. | ||
| * **Array Sizes:** nums has length 2-500, and we have 1-500 range queries. Each query specifies a subarray that can be rearranged. | ||
| * **Time Complexity:** For each query, we extract the subarray (O(n)), sort it (O(n log n)), and check if it's arithmetic (O(n)). With m queries, total is O(m * n log n). **Time Complexity: O(m * n log n)**, **Space Complexity: O(n)** for the sorted subarray. | ||
| * **Edge Case:** If a subarray has less than 2 elements, it's trivially arithmetic (return True). | ||
|
|
||
| **1.2 High-level approach** | ||
| Extract subarray, sort it, ensure consecutive differences are equal. | ||
|  | ||
| **High-level approach** | ||
|
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||
| **1.3 Brute force vs. optimized strategy** | ||
| - **Brute Force:** Try all permutations — factorial blowup. | ||
| - **Optimized:** Sorting then one pass diff check — O(k log k). | ||
| The problem asks us to check if subarrays can be rearranged to form arithmetic sequences. A sequence is arithmetic if the difference between consecutive elements is constant. | ||
|
|
||
| **1.4 Decomposition** | ||
| 1. For each query, copy subarray `nums[l:r+1]`. | ||
| 2. Sort it. | ||
| 3. Compute common diff = `sorted[1]-sorted[0]`. | ||
| 4. Verify all consecutive diffs match; if any differ, mark false; else true. | ||
| **Brute force vs. optimized strategy** | ||
|
|
||
| * **Brute Force:** For each query, try all permutations of the subarray to see if any forms an arithmetic sequence. This would be O(n! * n) per query, which is exponential. | ||
| * **Optimized:** Sort the subarray and check if the sorted version is arithmetic. If a set of numbers can form an arithmetic sequence, its sorted version will be arithmetic. This is O(n log n) per query. | ||
|
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||
| **Decomposition** | ||
|
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||
| 1. **Extract Subarray:** Get the subarray from nums[l[i]] to nums[r[i]]. | ||
| 2. **Sort:** Sort the subarray to check if it can be arithmetic. | ||
| 3. **Check Arithmetic:** Verify that differences between consecutive elements are constant. | ||
|
|
||
| ### Steps | ||
|
|
||
| **2.1 Initialization & Example Setup** | ||
| Example: `nums=[4,6,5,9,3,7]`, query `[0,2]` → subarray `[4,6,5]`. | ||
| 1. **Initialization & Example Setup** | ||
| Let's use `nums = [4,6,5,9,3,7]`, `l = [0,0,2]`, `r = [2,3,5]` as our example. | ||
| - Initialize `res = []` to store results. | ||
|
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||
| **2.2 Start Checking** | ||
| Sort → `[4,5,6]`, diff = 1. | ||
| 2. **Process Each Query** | ||
| For query i = 0: `l[0]=0`, `r[0]=2` | ||
| - Extract subarray: `nums[0:3] = [4,6,5]` | ||
| - Sort: `[4,5,6]` | ||
| - Check differences: `5-4=1`, `6-5=1` → constant difference → True | ||
|
|
||
| **2.3 Trace Walkthrough** | ||
| | Subarray (sorted) | Step | Diff check | OK? | | ||
| |-------------------|------|----------------------|-----| | ||
| | [4,5,6] | 4→5 | 1 == diff | ✓ | | ||
| | | 5→6 | 1 == diff | ✓ | | ||
| 3. **Trace Walkthrough** | ||
|
|
||
| **2.4 Increment and Loop** | ||
| Repeat for each query independently. | ||
| | Query | l[i] | r[i] | Subarray | Sorted | Differences | Result | | ||
| |-------|------|------|----------|--------|-------------|--------| | ||
| | 0 | 0 | 2 | [4,6,5] | [4,5,6] | 1, 1 | True | | ||
| | 1 | 0 | 3 | [4,6,5,9]| [4,5,6,9]| 1, 1, 3 | False | | ||
| | 2 | 2 | 5 | [5,9,3,7]| [3,5,7,9]| 2, 2, 2 | True | | ||
|
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||
| **2.5 Return Result** | ||
| Append boolean for each query into `res`. | ||
| 4. **Check Arithmetic Logic** | ||
| - If sorted array has less than 2 elements, return True. | ||
| - Calculate `diff = sorted[1] - sorted[0]`. | ||
| - For each subsequent pair, check if `sorted[i] - sorted[i-1] == diff`. | ||
|
|
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🛠️ Refactor suggestion | 🟠 Major
Use Markdown headings (###) instead of bold emphasis for section headers.
All 8 explanation files use bold text (e.g., **Constraints & Edge Cases**) for subsection headers, which triggers markdownlint rule MD036. Markdown processors and tools expect semantic heading hierarchy. Replace bold emphasis with ### (or #### if nesting under ### Strategy).
For all 8 files (1630, 2326, 1551, 2221, 1261, 797, 1605, 2120), apply this pattern to lines 5, 11, 15, 20:
- **Constraints & Edge Cases**
+ ### Constraints & Edge Cases
- **High-level approach**
+ ### High-level approach
- **Brute force vs. optimized strategy**
+ ### Brute force vs. optimized strategy
- **Decomposition**
+ ### DecompositionThen adjust subsequent content (lines 7–50) to indent one level (add one space or adjust list indentation) if they were under ### Strategy context.
📝 Committable suggestion
‼️ IMPORTANT
Carefully review the code before committing. Ensure that it accurately replaces the highlighted code, contains no missing lines, and has no issues with indentation. Thoroughly test & benchmark the code to ensure it meets the requirements.
| **Constraints & Edge Cases** | |
| **1.1 Constraints & Complexity** | |
| - **Input Size:** `n, m <= 500`. | |
| - **Time Complexity:** O(m * k log k), where k is subarray length (sort each subarray). | |
| - **Space Complexity:** O(k) per query for the sorted copy. | |
| - **Edge Case:** Subarray of length 2 is always arithmetic. | |
| * **Array Sizes:** nums has length 2-500, and we have 1-500 range queries. Each query specifies a subarray that can be rearranged. | |
| * **Time Complexity:** For each query, we extract the subarray (O(n)), sort it (O(n log n)), and check if it's arithmetic (O(n)). With m queries, total is O(m * n log n). **Time Complexity: O(m * n log n)**, **Space Complexity: O(n)** for the sorted subarray. | |
| * **Edge Case:** If a subarray has less than 2 elements, it's trivially arithmetic (return True). | |
| **1.2 High-level approach** | |
| Extract subarray, sort it, ensure consecutive differences are equal. | |
|  | |
| **High-level approach** | |
| **1.3 Brute force vs. optimized strategy** | |
| - **Brute Force:** Try all permutations — factorial blowup. | |
| - **Optimized:** Sorting then one pass diff check — O(k log k). | |
| The problem asks us to check if subarrays can be rearranged to form arithmetic sequences. A sequence is arithmetic if the difference between consecutive elements is constant. | |
| **1.4 Decomposition** | |
| 1. For each query, copy subarray `nums[l:r+1]`. | |
| 2. Sort it. | |
| 3. Compute common diff = `sorted[1]-sorted[0]`. | |
| 4. Verify all consecutive diffs match; if any differ, mark false; else true. | |
| **Brute force vs. optimized strategy** | |
| * **Brute Force:** For each query, try all permutations of the subarray to see if any forms an arithmetic sequence. This would be O(n! * n) per query, which is exponential. | |
| * **Optimized:** Sort the subarray and check if the sorted version is arithmetic. If a set of numbers can form an arithmetic sequence, its sorted version will be arithmetic. This is O(n log n) per query. | |
| **Decomposition** | |
| 1. **Extract Subarray:** Get the subarray from nums[l[i]] to nums[r[i]]. | |
| 2. **Sort:** Sort the subarray to check if it can be arithmetic. | |
| 3. **Check Arithmetic:** Verify that differences between consecutive elements are constant. | |
| ### Steps | |
| **2.1 Initialization & Example Setup** | |
| Example: `nums=[4,6,5,9,3,7]`, query `[0,2]` → subarray `[4,6,5]`. | |
| 1. **Initialization & Example Setup** | |
| Let's use `nums = [4,6,5,9,3,7]`, `l = [0,0,2]`, `r = [2,3,5]` as our example. | |
| - Initialize `res = []` to store results. | |
| **2.2 Start Checking** | |
| Sort → `[4,5,6]`, diff = 1. | |
| 2. **Process Each Query** | |
| For query i = 0: `l[0]=0`, `r[0]=2` | |
| - Extract subarray: `nums[0:3] = [4,6,5]` | |
| - Sort: `[4,5,6]` | |
| - Check differences: `5-4=1`, `6-5=1` → constant difference → True | |
| **2.3 Trace Walkthrough** | |
| | Subarray (sorted) | Step | Diff check | OK? | | |
| |-------------------|------|----------------------|-----| | |
| | [4,5,6] | 4→5 | 1 == diff | ✓ | | |
| | | 5→6 | 1 == diff | ✓ | | |
| 3. **Trace Walkthrough** | |
| **2.4 Increment and Loop** | |
| Repeat for each query independently. | |
| | Query | l[i] | r[i] | Subarray | Sorted | Differences | Result | | |
| |-------|------|------|----------|--------|-------------|--------| | |
| | 0 | 0 | 2 | [4,6,5] | [4,5,6] | 1, 1 | True | | |
| | 1 | 0 | 3 | [4,6,5,9]| [4,5,6,9]| 1, 1, 3 | False | | |
| | 2 | 2 | 5 | [5,9,3,7]| [3,5,7,9]| 2, 2, 2 | True | | |
| **2.5 Return Result** | |
| Append boolean for each query into `res`. | |
| 4. **Check Arithmetic Logic** | |
| - If sorted array has less than 2 elements, return True. | |
| - Calculate `diff = sorted[1] - sorted[0]`. | |
| - For each subsequent pair, check if `sorted[i] - sorted[i-1] == diff`. | |
| ### Constraints & Edge Cases | |
| * **Array Sizes:** nums has length 2-500, and we have 1-500 range queries. Each query specifies a subarray that can be rearranged. | |
| * **Time Complexity:** For each query, we extract the subarray (O(n)), sort it (O(n log n)), and check if it's arithmetic (O(n)). With m queries, total is O(m * n log n). **Time Complexity: O(m * n log n)**, **Space Complexity: O(n)** for the sorted subarray. | |
| * **Edge Case:** If a subarray has less than 2 elements, it's trivially arithmetic (return True). | |
| ### High-level approach | |
| The problem asks us to check if subarrays can be rearranged to form arithmetic sequences. A sequence is arithmetic if the difference between consecutive elements is constant. | |
| ### Brute force vs. optimized strategy | |
| * **Brute Force:** For each query, try all permutations of the subarray to see if any forms an arithmetic sequence. This would be O(n! * n) per query, which is exponential. | |
| * **Optimized:** Sort the subarray and check if the sorted version is arithmetic. If a set of numbers can form an arithmetic sequence, its sorted version will be arithmetic. This is O(n log n) per query. | |
| ### Decomposition | |
| 1. **Extract Subarray:** Get the subarray from nums[l[i]] to nums[r[i]]. | |
| 2. **Sort:** Sort the subarray to check if it can be arithmetic. | |
| 3. **Check Arithmetic:** Verify that differences between consecutive elements are constant. | |
| ### Steps | |
| 1. **Initialization & Example Setup** | |
| Let's use `nums = [4,6,5,9,3,7]`, `l = [0,0,2]`, `r = [2,3,5]` as our example. | |
| - Initialize `res = []` to store results. | |
| 2. **Process Each Query** | |
| For query i = 0: `l[0]=0`, `r[0]=2` | |
| - Extract subarray: `nums[0:3] = [4,6,5]` | |
| - Sort: `[4,5,6]` | |
| - Check differences: `5-4=1`, `6-5=1` → constant difference → True | |
| 3. **Trace Walkthrough** | |
| | Query | l[i] | r[i] | Subarray | Sorted | Differences | Result | | |
| |-------|------|------|----------|--------|-------------|--------| | |
| | 0 | 0 | 2 | [4,6,5] | [4,5,6] | 1, 1 | True | | |
| | 1 | 0 | 3 | [4,6,5,9]| [4,5,6,9]| 1, 1, 3 | False | | |
| | 2 | 2 | 5 | [5,9,3,7]| [3,5,7,9]| 2, 2, 2 | True | | |
| 4. **Check Arithmetic Logic** | |
| - If sorted array has less than 2 elements, return True. | |
| - Calculate `diff = sorted[1] - sorted[0]`. | |
| - For each subsequent pair, check if `sorted[i] - sorted[i-1] == diff`. |
🧰 Tools
🪛 markdownlint-cli2 (0.18.1)
5-5: Emphasis used instead of a heading
(MD036, no-emphasis-as-heading)
11-11: Emphasis used instead of a heading
(MD036, no-emphasis-as-heading)
15-15: Emphasis used instead of a heading
(MD036, no-emphasis-as-heading)
20-20: Emphasis used instead of a heading
(MD036, no-emphasis-as-heading)
| - If we buy employee 0: child gets discount → `dfs(1, True)` returns `{1: 1}` (cost=1, profit=3-1=2, but wait...) | ||
| - Merge: `buy_current` with child discount results |
There was a problem hiding this comment.
Incomplete explanation comment.
Line 40 contains "but wait..." which appears to be an incomplete thought left in the documentation. Consider either completing the explanation or removing this parenthetical remark for clarity.
- - If we buy employee 0: child gets discount → `dfs(1, True)` returns `{1: 1}` (cost=1, profit=3-1=2, but wait...)
+ - If we buy employee 0: child gets discount → `dfs(1, True)` returns `{1: 2}` (discounted cost=2//2=1, profit=3-1=2)📝 Committable suggestion
‼️ IMPORTANT
Carefully review the code before committing. Ensure that it accurately replaces the highlighted code, contains no missing lines, and has no issues with indentation. Thoroughly test & benchmark the code to ensure it meets the requirements.
| - If we buy employee 0: child gets discount → `dfs(1, True)` returns `{1: 1}` (cost=1, profit=3-1=2, but wait...) | |
| - Merge: `buy_current` with child discount results | |
| - If we buy employee 0: child gets discount → `dfs(1, True)` returns `{1: 2}` (discounted cost=2//2=1, profit=3-1=2) | |
| - Merge: `buy_current` with child discount results |
🤖 Prompt for AI Agents
In explanations/3562/en.md around lines 40 to 41, the sentence fragment "but
wait..." is an incomplete parenthetical that breaks the explanation; either
complete the thought by explicitly stating the discrepancy (e.g., explain why
cost vs. profit calculation needs adjustment or which values are being compared)
or simply remove the "but wait..." fragment so the example reads cleanly and the
merge step flows without an unfinished remark.
| from typing import List | ||
|
|
||
| class Solution: | ||
| def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]: |
There was a problem hiding this comment.
Missing List import for type annotations.
Same issue as other files - List is used but not imported.
+from typing import List
+
class Solution:
def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]:🧰 Tools
🪛 Ruff (0.14.8)
2-2: Undefined name List
(F821)
2-2: Ambiguous variable name: l
(E741)
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
🤖 Prompt for AI Agents
In solutions/1630/01.py around line 2, the function signature uses the typing
name List but the module does not import List; add the missing import from the
typing module (e.g., from typing import List) at the top of the file so the type
annotations resolve correctly.
| def maxProfit(self, n: int, present: List[int], future: List[int], hierarchy: List[List[int]], budget: int) -> int: | ||
| from collections import defaultdict | ||
| from functools import lru_cache |
There was a problem hiding this comment.
Missing List import from typing.
The type hint List[int] is used but List is not imported, which will cause a NameError at runtime.
+from typing import List
+
class Solution:
def maxProfit(self, n: int, present: List[int], future: List[int], hierarchy: List[List[int]], budget: int) -> int:
from collections import defaultdict
from functools import lru_cacheAlternatively, for Python 3.9+, use the built-in list lowercase:
def maxProfit(self, n: int, present: list[int], future: list[int], hierarchy: list[list[int]], budget: int) -> int:🧰 Tools
🪛 Ruff (0.14.8)
2-2: Unused method argument: n
(ARG002)
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
🤖 Prompt for AI Agents
In solutions/3562/01.py around lines 2 to 4, the function signature uses the
type name List but List is not imported; add a typing import or switch to
built-in generics to fix the NameError. Either add "from typing import List" at
the top (or alongside the existing imports) or update the annotations to use
Python 3.9+ built-in types (e.g., list[int], list[list[int]]) so the type names
resolve correctly.
romankurnovskii
deleted the
problems-3562-2482-1261-1008-1630-1079-2785-797-237-3271-2637-1605-1551-1347-2326-1823-2221-2120
branch
2 participants
Summary
This PR adds solutions and explanations for 16 LeetCode problems.
Accepted Problems (16)
237 - Delete Node in a Linked List (Medium)
797 - All Paths From Source to Target (Medium)
1008 - Construct Binary Search Tree from Preorder Traversal (Medium)
1079 - Letter Tile Possibilities (Medium)
1261 - Find Elements in a Contaminated Binary Tree (Medium)
1347 - Minimum Number of Steps to Make Two Strings Anagram (Medium)
1551 - Minimum Operations to Make Array Equal (Medium)
1605 - Find Valid Matrix Given Row and Column Sums (Medium)
1630 - Arithmetic Subarrays (Medium)
1823 - Find the Winner of the Circular Game (Medium)
2120 - Execution of All Suffix Instructions Staying in a Grid (Medium)
2221 - Find Triangular Sum of an Array (Medium)
2326 - Spiral Matrix IV (Medium)
2482 - Difference Between Ones and Zeros in Row and Column (Medium)
2785 - Sort Vowels in a String (Medium)
3562 - Maximum Profit from Trading Stocks with Discounts (Hard)
Notes
resas the result variable name.Summary by Sourcery
Add and adjust Python solutions and accompanying explanations for several LeetCode problems, including a new tree-DP/knapsack solution for problem 3562, while simplifying code and documentation structure.
New Features:
Bug Fixes:
Enhancements:
Summary by CodeRabbit
New Features
Documentation
Optimization
✏️ Tip: You can customize this high-level summary in your review settings.