Add solutions and explanations for problems 2390, 2542, 3753, 3759-3762, 3765-3772 #105
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WalkthroughThis PR adds eight new LeetCode problems (IDs 3765–3772) to the problem database, each with corresponding explanation documentation and Python solution implementations across diverse algorithm categories including Math, Arrays, Greedy, Dynamic Programming, and Trees. Changes
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Reviewer's GuideAdds Python solutions and English explanations for multiple LeetCode problems (2390, 2542, 3765-3772, 3766-3769) plus small consistency tweaks to an existing 3762 explanation and solution, implementing standard patterns (stack, heap, DP, graph, sieve, Fenwick tree, tree DP) with result stored in Class diagram for FenwickTree-based minimum inversion count (problem 3768)classDiagram
class Solution {
+minInversionCount(nums List~int~, k int) int
}
class FenwickTree {
-n int
-tree List~int~
+__init__(size int)
+update(idx int, delta int) void
+query(idx int) int
}
Solution --> FenwickTree : uses
Flow diagram for heap-based maximum subsequence score (problem 2542)graph TD
A["Start: nums1, nums2, k"] --> B["Create pairs (nums1[i], nums2[i])"]
B --> C["Sort pairs by nums2 descending"]
C --> D["Initialize min heap, current_sum = 0, res = 0"]
D --> E{More pairs?}
E -->|Yes| F["Take next pair (num1, num2)"]
F --> G["Push num1 into heap and add to current_sum"]
G --> H{heap size > k?}
H -->|Yes| I["Pop smallest from heap and subtract from current_sum"]
H -->|No| J["Skip pop"]
I --> K{heap size == k?}
J --> K{heap size == k?}
K -->|Yes| L["score = current_sum * num2"]
L --> M["res = max(res, score)"]
K -->|No| N["Continue"]
M --> E
N --> E
E -->|No| O["Return res"]
Flow diagram for stack-based star removal in string (problem 2390)graph TD
A["Start: input string s"] --> B["Initialize empty stack"]
B --> C{More characters in s?}
C -->|Yes| D["Read next char"]
D --> E{char == '*'?}
E -->|Yes| F{stack not empty?}
F -->|Yes| G["Pop top of stack"]
F -->|No| H["Do nothing"]
E -->|No| I["Push char onto stack"]
G --> C
H --> C
I --> C
C -->|No| J["Join stack into string res"]
J --> K["Return res"]
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Hey there - I've reviewed your changes and found some issues that need to be addressed.
- In
solutions/3768/01.pythe sliding window inversion update for the leftmost element looks incorrect: you callft.update(compressed[i - k], -1)before computing how many smaller elements it dominates and then subtractsmallerfrominv_count; you should query the tree (with the element still present) to get the number of elements it forms inversions with, then remove it, otherwise you’re subtracting the wrong quantity. - The implementation for
maxSubgraphScoreinsolutions/3772/01.pyruns a full DFS from every root (O(n^2) on a tree of size up to 1e5), which is unlikely to scale; consider a standard two-pass tree DP/rerooting approach that computes all answers in O(n). - Both
solutions/3771/01.py(O(n^2) double loop) andsolutions/3770/01.py(nested loops over primes with cumulative sums) have worst-case runtimes that are too high for the stated constraints (n up to 1e5 and 5e5 respectively); it would be good to revisit these with more scalable strategies (e.g., leveraging prefix sums and pruning/early stopping or more targeted math/DP optimizations).
Prompt for AI Agents
Please address the comments from this code review:
## Overall Comments
- In `solutions/3768/01.py` the sliding window inversion update for the leftmost element looks incorrect: you call `ft.update(compressed[i - k], -1)` before computing how many smaller elements it dominates and then subtract `smaller` from `inv_count`; you should query the tree (with the element still present) to get the number of elements it forms inversions with, then remove it, otherwise you’re subtracting the wrong quantity.
- The implementation for `maxSubgraphScore` in `solutions/3772/01.py` runs a full DFS from every root (O(n^2) on a tree of size up to 1e5), which is unlikely to scale; consider a standard two-pass tree DP/rerooting approach that computes all answers in O(n).
- Both `solutions/3771/01.py` (O(n^2) double loop) and `solutions/3770/01.py` (nested loops over primes with cumulative sums) have worst-case runtimes that are too high for the stated constraints (n up to 1e5 and 5e5 respectively); it would be good to revisit these with more scalable strategies (e.g., leveraging prefix sums and pruning/early stopping or more targeted math/DP optimizations).
## Individual Comments
### Comment 1
<location> `solutions/3768/01.py:40-49` </location>
<code_context>
+ res = inv_count
+
+ # Slide window
+ for i in range(k, n):
+ # Remove leftmost element
+ ft.update(compressed[i - k], -1)
+ # Count numbers smaller than removed element
+ removed = compressed[i - k]
+ smaller = ft.query(removed - 1)
+ inv_count -= smaller
+
+ # Add new element
+ new_val = compressed[i]
+ # Count numbers greater than new element in current window
+ greater = (k - 1) - ft.query(new_val)
+ inv_count += greater
+ ft.update(new_val, 1)
+
+ res = min(res, inv_count)
</code_context>
<issue_to_address>
**issue (bug_risk):** The sliding-window inversion update logic looks incorrect and risks miscounting inversions.
The incremental updates to `inv_count` don’t match how inversions are defined over the window. When removing `removed`, subtracting `ft.query(removed - 1)` counts remaining elements smaller than `removed`, but the inversions involving `removed` depend on its relative position and which side of the pair it’s on. Likewise, `greater = (k - 1) - ft.query(new_val)` assumes a fixed window size and that `ft.query(new_val)` gives all elements ≤ `new_val`, but this doesn’t directly correspond to the number of new inversions. Please re-derive the sliding update formulas with respect to pair directions and positions, or, if feasible, recompute the inversion count per window with a fresh Fenwick tree for correctness.
</issue_to_address>
### Comment 2
<location> `solutions/3772/01.py:31-33` </location>
<code_context>
+
+ # For each node as root, calculate max score
+ res = []
+ for root in range(n):
+ score = dfs(root, -1)
+ res.append(score)
+
+ return res
</code_context>
<issue_to_address>
**suggestion (performance):** Running a DFS from every root leads to O(n²) time on a tree, which is unnecessary.
For trees, this can be done in O(n) or O(n log n) with a rerooting DP: one DFS to compute subtree scores, then a second DFS to propagate values when changing roots. Calling `dfs` for every `root` recomputes the same work O(n) times and is likely to TLE for large `n`. Consider restructuring to compute all root scores in a single rerooting pass instead.
Suggested implementation:
```python
# Rerooting DP: compute all root scores in O(n)
# 1) First DFS: compute subtree scores when rooted at an arbitrary node (e.g. 0)
subtree = [0] * n
def dfs_down(node: int, parent: int) -> int:
best = values[node]
for child in graph[node]:
if child != parent:
child_score = dfs_down(child, node)
if child_score > 0:
best += child_score
subtree[node] = best
return best
dfs_down(0, -1)
# 2) Second DFS: reroot, propagating contributions from the parent side
res = [0] * n
def dfs_reroot(node: int, parent: int, parent_contrib: int) -> None:
# Score when rooted at `node` = its subtree plus positive contribution from parent side
res[node] = subtree[node] + max(0, parent_contrib)
# Collect children (excluding parent) with their positive contributions
children = []
for child in graph[node]:
if child == parent:
continue
children.append((child, max(0, subtree[child])))
m = len(children)
prefix = [0] * (m + 1)
suffix = [0] * (m + 1)
for i in range(m):
prefix[i + 1] = prefix[i] + children[i][1]
suffix[m - 1 - i] = suffix[m - i] + children[m - 1 - i][1]
for i, (child, child_contrib) in enumerate(children):
# Contribution to pass to this child from "everything except this child"
without_child = values[node] + prefix[i] + suffix[i + 1]
if parent_contrib > 0:
without_child += parent_contrib
dfs_reroot(child, node, without_child)
dfs_reroot(0, -1, 0)
return res
```
This implementation assumes:
1. The node values are stored in an iterable named `values` indexed from `0` to `n-1`.
2. `graph` is an adjacency list for a tree with `n` nodes and is already built earlier in the function.
If your current `dfs` had a different initialization for `best` (e.g. starting from `0` instead of `values[node]` or incorporating other state), you should mirror that initialization in `dfs_down` to keep the semantics identical. Also remove or inline the original `dfs` if it is no longer used.
</issue_to_address>
### Comment 3
<location> `solutions/3771/01.py:6-15` </location>
<code_context>
+ n = len(damage)
+
+ # Prefix sums for damage
+ prefix = [0] * (n + 1)
+ for i in range(n):
+ prefix[i + 1] = prefix[i] + damage[i]
+
+ # For each starting position i, calculate score
+ res = 0
+ for i in range(n):
+ current_hp = hp
+ score = 0
+ for j in range(i, n):
+ current_hp -= damage[j]
+ if current_hp >= requirement[j]:
+ score += 1
+ res += score
+
+ return res
</code_context>
<issue_to_address>
**suggestion (performance):** The current O(n²) nested loops and unused prefix sum may be inefficient compared to prefix-based or two-pointer approaches.
The computed `prefix` array is unused; `current_hp` is still updated via a full inner loop (`current_hp -= damage[j]`). For large inputs this O(n²) pattern may be too slow. Consider using the prefix sums directly (e.g., `prefix[j + 1] - prefix[i]`) to enable a more efficient approach such as binary searching the maximal `j` per `i`, or otherwise reducing the complexity.
Suggested implementation:
```python
from bisect import bisect_left
class Solution:
def totalScore(self, hp: int, damage: List[int], requirement: List[int]) -> int:
n = len(damage)
# Prefix sums for damage
prefix = [0] * (n + 1)
for i in range(n):
prefix[i + 1] = prefix[i] + damage[i]
# Count valid (i, j) pairs using prefix sums and binary search.
# Condition for starting at i and ending at j:
# hp - (prefix[j + 1] - prefix[i]) >= requirement[j]
# => prefix[i] >= requirement[j] - hp + prefix[j + 1]
res = 0
for j in range(n):
threshold = requirement[j] - hp + prefix[j + 1]
# Find the first i in [0, j] such that prefix[i] >= threshold
idx = bisect_left(prefix, threshold, 0, j + 1)
if idx <= j:
res += (j - idx + 1)
return res
```
1. Ensure there is not already a conflicting `from bisect import bisect_left` import at the top of the file. If imports are grouped elsewhere, you may instead want to add `from bisect import bisect_left` alongside the existing imports rather than directly before `class Solution`.
2. This approach assumes that `damage` elements are non-negative so that `prefix` is non-decreasing (a typical assumption in HP/damage problems). If negative values are allowed, this monotonicity-based `bisect_left` logic must be revisited.
</issue_to_address>Help me be more useful! Please click 👍 or 👎 on each comment and I'll use the feedback to improve your reviews.
| for i in range(k, n): | ||
| # Remove leftmost element | ||
| ft.update(compressed[i - k], -1) | ||
| # Count numbers smaller than removed element | ||
| removed = compressed[i - k] | ||
| smaller = ft.query(removed - 1) | ||
| inv_count -= smaller | ||
|
|
||
| # Add new element | ||
| new_val = compressed[i] |
There was a problem hiding this comment.
issue (bug_risk): The sliding-window inversion update logic looks incorrect and risks miscounting inversions.
The incremental updates to inv_count don’t match how inversions are defined over the window. When removing removed, subtracting ft.query(removed - 1) counts remaining elements smaller than removed, but the inversions involving removed depend on its relative position and which side of the pair it’s on. Likewise, greater = (k - 1) - ft.query(new_val) assumes a fixed window size and that ft.query(new_val) gives all elements ≤ new_val, but this doesn’t directly correspond to the number of new inversions. Please re-derive the sliding update formulas with respect to pair directions and positions, or, if feasible, recompute the inversion count per window with a fresh Fenwick tree for correctness.
| for root in range(n): | ||
| score = dfs(root, -1) | ||
| res.append(score) |
There was a problem hiding this comment.
suggestion (performance): Running a DFS from every root leads to O(n²) time on a tree, which is unnecessary.
For trees, this can be done in O(n) or O(n log n) with a rerooting DP: one DFS to compute subtree scores, then a second DFS to propagate values when changing roots. Calling dfs for every root recomputes the same work O(n) times and is likely to TLE for large n. Consider restructuring to compute all root scores in a single rerooting pass instead.
Suggested implementation:
# Rerooting DP: compute all root scores in O(n)
# 1) First DFS: compute subtree scores when rooted at an arbitrary node (e.g. 0)
subtree = [0] * n
def dfs_down(node: int, parent: int) -> int:
best = values[node]
for child in graph[node]:
if child != parent:
child_score = dfs_down(child, node)
if child_score > 0:
best += child_score
subtree[node] = best
return best
dfs_down(0, -1)
# 2) Second DFS: reroot, propagating contributions from the parent side
res = [0] * n
def dfs_reroot(node: int, parent: int, parent_contrib: int) -> None:
# Score when rooted at `node` = its subtree plus positive contribution from parent side
res[node] = subtree[node] + max(0, parent_contrib)
# Collect children (excluding parent) with their positive contributions
children = []
for child in graph[node]:
if child == parent:
continue
children.append((child, max(0, subtree[child])))
m = len(children)
prefix = [0] * (m + 1)
suffix = [0] * (m + 1)
for i in range(m):
prefix[i + 1] = prefix[i] + children[i][1]
suffix[m - 1 - i] = suffix[m - i] + children[m - 1 - i][1]
for i, (child, child_contrib) in enumerate(children):
# Contribution to pass to this child from "everything except this child"
without_child = values[node] + prefix[i] + suffix[i + 1]
if parent_contrib > 0:
without_child += parent_contrib
dfs_reroot(child, node, without_child)
dfs_reroot(0, -1, 0)
return resThis implementation assumes:
- The node values are stored in an iterable named
valuesindexed from0ton-1. graphis an adjacency list for a tree withnnodes and is already built earlier in the function.
If your current dfs had a different initialization for best (e.g. starting from 0 instead of values[node] or incorporating other state), you should mirror that initialization in dfs_down to keep the semantics identical. Also remove or inline the original dfs if it is no longer used.
| prefix = [0] * (n + 1) | ||
| for i in range(n): | ||
| prefix[i + 1] = prefix[i] + damage[i] | ||
|
|
||
| # For each starting position i, calculate score | ||
| res = 0 | ||
| for i in range(n): | ||
| current_hp = hp | ||
| score = 0 | ||
| for j in range(i, n): |
There was a problem hiding this comment.
suggestion (performance): The current O(n²) nested loops and unused prefix sum may be inefficient compared to prefix-based or two-pointer approaches.
The computed prefix array is unused; current_hp is still updated via a full inner loop (current_hp -= damage[j]). For large inputs this O(n²) pattern may be too slow. Consider using the prefix sums directly (e.g., prefix[j + 1] - prefix[i]) to enable a more efficient approach such as binary searching the maximal j per i, or otherwise reducing the complexity.
Suggested implementation:
from bisect import bisect_left
class Solution:
def totalScore(self, hp: int, damage: List[int], requirement: List[int]) -> int:
n = len(damage)
# Prefix sums for damage
prefix = [0] * (n + 1)
for i in range(n):
prefix[i + 1] = prefix[i] + damage[i]
# Count valid (i, j) pairs using prefix sums and binary search.
# Condition for starting at i and ending at j:
# hp - (prefix[j + 1] - prefix[i]) >= requirement[j]
# => prefix[i] >= requirement[j] - hp + prefix[j + 1]
res = 0
for j in range(n):
threshold = requirement[j] - hp + prefix[j + 1]
# Find the first i in [0, j] such that prefix[i] >= threshold
idx = bisect_left(prefix, threshold, 0, j + 1)
if idx <= j:
res += (j - idx + 1)
return res- Ensure there is not already a conflicting
from bisect import bisect_leftimport at the top of the file. If imports are grouped elsewhere, you may instead want to addfrom bisect import bisect_leftalongside the existing imports rather than directly beforeclass Solution. - This approach assumes that
damageelements are non-negative so thatprefixis non-decreasing (a typical assumption in HP/damage problems). If negative values are allowed, this monotonicity-basedbisect_leftlogic must be revisited.
romankurnovskii
deleted the
problems-2390-2542-3753-3759-3760-3761-3762-3765-3766-3767-3768-3769-3770-3771-3772
branch
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Actionable comments posted: 9
Caution
Some comments are outside the diff and can’t be posted inline due to platform limitations.
⚠️ Outside diff range comments (1)
solutions/3762/01.py (1)
1-14: Add missingListimport from typingLine 2 uses
List[int]in the type hints, but there is no import statement visible in the provided code. UsingList[int]without importingListfromtypingwill raiseNameError: name 'List' is not definedat runtime on Python versions before 3.9.Add the import:
+from typing import List + class Solution: def minOperations(self, nums: List[int], k: int) -> int:Alternatively, use the built-in generic syntax on Python 3.9+:
list[int]instead ofList[int], or addfrom __future__ import annotationsat the top for forward compatibility.
🧹 Nitpick comments (1)
solutions/3767/01.py (1)
1-23: Consider early break in the greedy loop for efficiencyThe greedy algorithm is correct: sorting deltas in descending order and applying the largest switches maximizes points. However, since
deltasis sorted descending, once a non-positive delta is encountered, further iterations cannot improvetotal. You can optionally break early to skip unnecessary work:for i in range(n - k): + if deltas[i] <= 0: + break total += deltas[i] res = max(res, total)This optimization saves iterations when many negative or zero deltas exist, though it doesn't affect correctness.
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📒 Files selected for processing (23)
data/leetcode-problems.json(1 hunks)explanations/2390/en.md(1 hunks)explanations/2542/en.md(1 hunks)explanations/3762/en.md(1 hunks)explanations/3765/en.md(1 hunks)explanations/3766/en.md(1 hunks)explanations/3767/en.md(1 hunks)explanations/3768/en.md(1 hunks)explanations/3769/en.md(1 hunks)explanations/3770/en.md(1 hunks)explanations/3771/en.md(1 hunks)explanations/3772/en.md(1 hunks)solutions/2390/01.py(1 hunks)solutions/2542/01.py(1 hunks)solutions/3762/01.py(1 hunks)solutions/3765/01.py(1 hunks)solutions/3766/01.py(1 hunks)solutions/3767/01.py(1 hunks)solutions/3768/01.py(1 hunks)solutions/3769/01.py(1 hunks)solutions/3770/01.py(1 hunks)solutions/3771/01.py(1 hunks)solutions/3772/01.py(1 hunks)
🧰 Additional context used
🧬 Code graph analysis (1)
solutions/3767/01.py (1)
solutions/2542/01.py (1)
Solution(1-27)
🪛 GitHub Actions: PR JSON Validation
data/leetcode-problems.json
[error] 1-1: JSON file is not normalized. Run './scripts/normalize_json.py data/leetcode-problems.json' and commit the changes.
[warning] 26416-26416: No newline at end of file detected in data/leetcode-problems.json.
🪛 LanguageTool
explanations/3770/en.md
[duplication] ~9-~9: Possible typo: you repeated a word.
Context: ...<= 5 * 10^5` - Time Complexity: O(n log log n) for sieve + O(p²) for consecutive su...
(ENGLISH_WORD_REPEAT_RULE)
[duplication] ~20-~20: Possible typo: you repeated a word.
Context: ... Eratosthenes to mark all primes in O(n log log n), then compute consecutive sums - **E...
(ENGLISH_WORD_REPEAT_RULE)
explanations/3771/en.md
[style] ~11-~11: As an alternative to the over-used intensifier ‘very’, consider replacing this phrase.
Context: ...efix sums - Edge Case: If damage is very large, HP may become negative quickly **1.2 ...
(EN_WEAK_ADJECTIVE)
🪛 markdownlint-cli2 (0.18.1)
explanations/3766/en.md
42-42: Table column count
Expected: 5; Actual: 7; Too many cells, extra data will be missing
(MD056, table-column-count)
43-43: Table column count
Expected: 5; Actual: 7; Too many cells, extra data will be missing
(MD056, table-column-count)
44-44: Table column count
Expected: 5; Actual: 7; Too many cells, extra data will be missing
(MD056, table-column-count)
🪛 Ruff (0.14.7)
solutions/3768/01.py
2-2: Undefined name List
(F821)
solutions/3767/01.py
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
solutions/3766/01.py
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
solutions/3772/01.py
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
solutions/2542/01.py
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
4-4: zip() without an explicit strict= parameter
Add explicit value for parameter strict=
(B905)
solutions/3771/01.py
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
solutions/3769/01.py
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
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🔇 Additional comments (9)
solutions/3765/01.py (1)
1-30: Complete-prime implementation is correct and efficientThe
is_primehelper and the prefix/suffix loops correctly enforce that every prefix and suffix ofnumis prime, with reasonable √n primality checks and early returns. No issues spotted.explanations/3769/en.md (1)
1-50: Binary-reflection explanation matches the implemented strategyThe write-up clearly defines binary reflection, states the correct O(n log n) complexity, and explains sorting by
(reflection, original)with a concrete trace table that matchessolutions/3769/01.py. Looks consistent and ready to ship.solutions/3770/01.py (1)
1-27: Consecutive-prime-sum sieve solution looks correctThe sieve construction, extraction of all primes ≤ n, and the double loop over start/end with
current_sum > nearly breaks correctly enumerate all consecutive-prime sums and track the largest one that is still prime and ≤ n. No correctness or maintainability issues detected.solutions/3769/01.py (1)
1-11: AddListimport for type hints; reflection-based sorting logic is correctThe
sortByReflectionimplementation correctly computes binary reflections viabin(n)[2:], reversal, andint(..., 2), then sorts by(binary_reflection(x), x)to handle tie-breaking. However, the code usesList[int]without an import statement visible in lines 1-11. Depending on Python version and whetherfrom __future__ import annotationsis present elsewhere in the file, this may cause aNameError. Add the typing import:+from typing import List + class Solution: def sortByReflection(self, nums: List[int]) -> List[int]:solutions/2390/01.py (1)
1-16: Stack-based implementation is correct and efficientThe logic cleanly matches the problem specification (push non-
*, pop on*) with O(n) time and O(n) space and a clearresresult variable. No issues from a correctness or style standpoint.explanations/2390/en.md (1)
1-59: Explanation aligns well with the implemented solutionProblem restatement, complexity, stack strategy, and the step-by-step trace are all consistent with the Python implementation and standard solutions for 2390.
explanations/3768/en.md (1)
1-52: Clear and accurate description of the Fenwick + sliding-window approachThe explanation’s constraints, complexity, decomposition, and example all align with the implemented algorithm in
solutions/3768/01.py, and it’s technically sound.data/leetcode-problems.json (2)
26360-26414: Data entries are well-structured and consistent.The 8 new problem entries (3765–3772) follow the established schema with all required fields (
id,category,title,difficulty,link) properly populated. Categories align with problem domains, difficulty levels are appropriate, and LeetCode URLs follow the expected pattern. Content looks good once the JSON normalization and EOF newline issues are resolved.
26360-26414: The file is already properly formatted and does not require the suggested actions.The JSON file already has:
- Keys in numeric order
- Proper 2-space indentation
- A newline at EOF
Running the normalization script is unnecessary; the file meets all formatting requirements as-is. The problem entries (3765-3772) are correctly structured and positioned.
Likely an incorrect or invalid review comment.
| | Number | Binary | Closest Palindrome | Operations | Result | | ||
| |--------|--------|-------------------|------------|--------| | ||
| | 1 | 1 | 1 | |0| = 0 | 0 | | ||
| | 2 | 10 | 3 (11) | |2-3| = 1 | 1 | | ||
| | 4 | 100 | 3 (11) | |4-3| = 1 | 1 | | ||
|
|
There was a problem hiding this comment.
Fix table formatting and clarify operations expressions
The absolute-value notation in the Operations column introduces extra | characters, breaking the table (and triggering MD056), and the first row’s formula is a bit unclear.
You can avoid both issues by rewriting the Operations cells without | characters:
-| 1 | 1 | 1 | |0| = 0 | 0 |
-| 2 | 10 | 3 (11) | |2-3| = 1 | 1 |
-| 4 | 100 | 3 (11) | |4-3| = 1 | 1 |
+| 1 | 1 | 1 | abs(1-1) = 0 | 0 |
+| 2 | 10 | 3 (11) | abs(2-3) = 1 | 1 |
+| 4 | 100 | 3 (11) | abs(4-3) = 1 | 1 |This keeps the table well-formed and makes the math explicit.
🧰 Tools
🪛 markdownlint-cli2 (0.18.1)
42-42: Table column count
Expected: 5; Actual: 7; Too many cells, extra data will be missing
(MD056, table-column-count)
43-43: Table column count
Expected: 5; Actual: 7; Too many cells, extra data will be missing
(MD056, table-column-count)
44-44: Table column count
Expected: 5; Actual: 7; Too many cells, extra data will be missing
(MD056, table-column-count)
🤖 Prompt for AI Agents
In explanations/3766/en.md around lines 40–45, the Operations column uses
literal '|' characters for absolute-value notation which breaks the Markdown
table; replace those with a plain, pipe-free expression (e.g., use abs(1-1) = 0,
abs(2-3) = 1, abs(4-3) = 1) and update the first row to explicitly show abs(1-1)
= 0 so the table cells contain no '|' characters and render correctly.
| **2.1 Initialization & Example Setup:** | ||
| Let's use the example: `technique1 = [5,2,10]`, `technique2 = [10,3,8]`, `k = 2` | ||
| - Initial total: 5 + 2 + 10 = 17 | ||
| - Deltas: [10-5, 3-2, 8-10] = [5, 1, -2] | ||
|
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||
| **2.2 Start Processing:** | ||
| We sort deltas and apply the best switches. | ||
|
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||
| **2.3 Trace Walkthrough:** | ||
|
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||
| | Step | Total | Delta Applied | New Total | Max | | ||
| |------|-------|---------------|-----------|-----| | ||
| | Initial | 17 | - | - | 17 | | ||
| | Apply delta 5 | 17 | +5 | 22 | 22 | | ||
| | Apply delta 1 | 22 | +1 | 23 | 23 | | ||
|
|
||
| Since k=2, we can switch at most 1 task (n-k=1). We apply the largest delta (5). | ||
|
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||
| **2.4 Increment and Loop:** | ||
| After applying the best (n-k) deltas, we have the maximum. | ||
|
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||
| **2.5 Return Result:** | ||
| The maximum total is 22 (switching task 0 from technique1 to technique2: 2 + 10 + 10 = 22). |
There was a problem hiding this comment.
Fix example walkthrough to respect the k-constraint.
For the example technique1 = [5,2,10], technique2 = [10,3,8], k = 2 (so n-k = 1), the table currently applies both deltas 5 and 1 (Lines 45–46), even though we’re allowed to switch at most one task away from technique1. Later you correctly state that we can switch at most one task.
To avoid confusion, adjust the trace so that only the best single delta (5) is applied, or explicitly mark the second row as an invalid step. Right now, the table and the textual explanation conflict.
🤖 Prompt for AI Agents
In explanations/3767/en.md around lines 32 to 54, the example walkthrough
violates the k-constraint by applying two deltas while k=2 (n-k=1): update the
trace so only the single largest delta (+5) is applied (show Initial -> Apply
delta 5 -> New Total 22 -> Max 22) and remove or mark the second delta
application row as invalid, and adjust the surrounding text to state explicitly
that only one task can be switched.
| **1.1 Constraints & Complexity:** | ||
| - Input size: `1 <= n <= 10^5`, `1 <= hp <= 10^9` | ||
| - **Time Complexity:** O(n²) for the straightforward approach, can be optimized with prefix sums | ||
| - **Space Complexity:** O(n) for prefix sums | ||
| - **Edge Case:** If damage is very large, HP may become negative quickly | ||
|
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||
| **1.2 High-level approach:** | ||
| For each starting position, simulate the journey through remaining rooms, tracking HP and counting points when HP >= requirement after taking damage. | ||
|
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||
|  | ||
|
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||
| **1.3 Brute force vs. optimized strategy:** | ||
| - **Brute Force:** For each starting position, simulate the entire journey, which is O(n²) | ||
| - **Optimized Strategy:** Use prefix sums to quickly calculate cumulative damage, but still need to check each room, achieving O(n²) in worst case | ||
| - **Emphasize the optimization:** Prefix sums help but we still need to check each room's requirement |
There was a problem hiding this comment.
Clarify O(n²) complexity against the stated n ≤ 10^5 constraint
With n up to 10^5 (Line 8), the described per-start simulation in O(n²) would do on the order of 10^10 iterations in the worst case, which is unlikely to be acceptable in Python. It would be good to clarify whether this is only a conceptual/brute-force description (and a faster solution is intended in code), or to outline a genuinely sub‑quadratic approach that matches these constraints.
🧰 Tools
🪛 LanguageTool
[style] ~11-~11: As an alternative to the over-used intensifier ‘very’, consider replacing this phrase.
Context: ...efix sums - Edge Case: If damage is very large, HP may become negative quickly **1.2 ...
(EN_WEAK_ADJECTIVE)
| **1.1 Constraints & Complexity:** | ||
| - Input size: `2 <= n <= 10^5` | ||
| - **Time Complexity:** O(n²) for naive approach, can be optimized with tree DP | ||
| - **Space Complexity:** O(n) for the tree and DP arrays | ||
| - **Edge Case:** If all nodes are bad, the best score for each node might be -1 | ||
|
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||
| **1.2 High-level approach:** | ||
| We use tree dynamic programming with rerooting. For each node as root, we perform DFS to find the maximum score of connected subgraphs containing that root. | ||
|
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||
|  | ||
|
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||
| **1.3 Brute force vs. optimized strategy:** | ||
| - **Brute Force:** For each node, try all possible connected subgraphs, which is exponential | ||
| - **Optimized Strategy:** Use tree DP where for each root, we calculate the best subtree score, achieving O(n²) time | ||
| - **Emphasize the optimization:** Tree structure allows efficient calculation of subtree scores |
There was a problem hiding this comment.
Resolve confusion around ‘naive’ vs ‘optimized’ complexity and scalability
Lines 8–10 say the naive time complexity is O(n²) and that it “can be optimized with tree DP,” but Section 1.3 still presents the tree-DP strategy as O(n²). This is confusing, and with n ≤ 10^5 a quadratic algorithm is borderline at best. Please clarify which approach is actually considered brute force (likely exponential over subgraphs) and what complexity your tree-DP solution targets (ideally O(n) with standard rerooting), or explicitly state if you’re intentionally presenting an O(n²) solution despite these constraints.
🤖 Prompt for AI Agents
In explanations/3772/en.md around lines 7–21, the complexity descriptions are
inconsistent: replace the claim that the “naive” approach is O(n²) with that
brute-force over all connected subgraphs is exponential, and update Section 1.3
to state the tree DP with standard rerooting runs in O(n) overall (or explicitly
state the achieved time complexity if your implementation differs), keeping
space O(n); make the wording consistent across Sections 1.1 and 1.3 and add a
short note if you intentionally present an O(n²) variant explaining why.
| class Solution: | ||
| def maxScore(self, nums1: List[int], nums2: List[int], k: int) -> int: | ||
| # Sort pairs by nums2 in descending order | ||
| pairs = sorted(zip(nums1, nums2), key=lambda x: x[1], reverse=True) | ||
|
|
||
| import heapq | ||
| # Use min heap to maintain k largest nums1 values |
There was a problem hiding this comment.
Import List and consider explicit strict on zip.
Listin the signature (Line 2) is undefined; this will raise at import time in this repo. Add a typing import, e.g.:
+from typing import List
+
class Solution:
def maxScore(self, nums1: List[int], nums2: List[int], k: int) -> int:
- # Sort pairs by nums2 in descending order
- pairs = sorted(zip(nums1, nums2), key=lambda x: x[1], reverse=True)
+ # Sort pairs by nums2 in descending order
+ pairs = sorted(
+ zip(nums1, nums2, strict=False),
+ key=lambda x: x[1],
+ reverse=True,
+ )(Keeping strict=False preserves current behavior while satisfying B905; if you want a hard check that the arrays have equal length, use strict=True instead.)
🧰 Tools
🪛 Ruff (0.14.7)
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
4-4: zip() without an explicit strict= parameter
Add explicit value for parameter strict=
(B905)
🤖 Prompt for AI Agents
In solutions/2542/01.py around lines 1 to 7, the type name List used in the
function signature is not imported and zip should be explicit about length
behavior; add "from typing import List" at the top of the file and change the
zip call to include strict=False (or strict=True if you want to enforce equal
lengths) so the sorted(zip(...), key=..., reverse=True) becomes
sorted(zip(nums1, nums2, strict=False), key=..., reverse=True).
| class Solution: | ||
| def minOperations(self, nums: List[int]) -> List[int]: | ||
| # Precompute binary palindromes up to 5000 | ||
| def is_binary_palindrome(n): | ||
| binary = bin(n)[2:] | ||
| return binary == binary[::-1] | ||
|
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||
| # Generate all binary palindromes | ||
| palindromes = [] | ||
| for i in range(1, 5001): | ||
| if is_binary_palindrome(i): | ||
| palindromes.append(i) | ||
|
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||
| res = [] | ||
| for num in nums: | ||
| # Find closest palindrome | ||
| min_ops = float('inf') | ||
| for pal in palindromes: | ||
| ops = abs(num - pal) | ||
| min_ops = min(min_ops, ops) | ||
| res.append(min_ops) | ||
|
|
||
| return res |
There was a problem hiding this comment.
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Length of output: 3838
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Add missing List import from typing
The type annotation on line 2 uses List[int] without importing it from the typing module. While this works in LeetCode's execution environment (which auto-provides typing imports), it violates Python's import conventions and would cause a NameError if run locally.
Add the import at the top of the file:
+from typing import List
+
class Solution:
def minOperations(self, nums: List[int]) -> List[int]:This same pattern appears in other files across the repository (e.g., solutions/1029/01.py, solutions/1046/01.py), so consider standardizing imports across all solution files.
📝 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.
| class Solution: | |
| def minOperations(self, nums: List[int]) -> List[int]: | |
| # Precompute binary palindromes up to 5000 | |
| def is_binary_palindrome(n): | |
| binary = bin(n)[2:] | |
| return binary == binary[::-1] | |
| # Generate all binary palindromes | |
| palindromes = [] | |
| for i in range(1, 5001): | |
| if is_binary_palindrome(i): | |
| palindromes.append(i) | |
| res = [] | |
| for num in nums: | |
| # Find closest palindrome | |
| min_ops = float('inf') | |
| for pal in palindromes: | |
| ops = abs(num - pal) | |
| min_ops = min(min_ops, ops) | |
| res.append(min_ops) | |
| return res | |
| from typing import List | |
| class Solution: | |
| def minOperations(self, nums: List[int]) -> List[int]: | |
| # Precompute binary palindromes up to 5000 | |
| def is_binary_palindrome(n): | |
| binary = bin(n)[2:] | |
| return binary == binary[::-1] | |
| # Generate all binary palindromes | |
| palindromes = [] | |
| for i in range(1, 5001): | |
| if is_binary_palindrome(i): | |
| palindromes.append(i) | |
| res = [] | |
| for num in nums: | |
| # Find closest palindrome | |
| min_ops = float('inf') | |
| for pal in palindromes: | |
| ops = abs(num - pal) | |
| min_ops = min(min_ops, ops) | |
| res.append(min_ops) | |
| return res |
🧰 Tools
🪛 Ruff (0.14.7)
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
🤖 Prompt for AI Agents
In solutions/3766/01.py around lines 1 to 23, the type annotation List[int] is
used but List is not imported; add the missing import "from typing import List"
at the top of the file so the annotation resolves, and apply the same import to
other solution files that use List (e.g., solutions/1029/01.py,
solutions/1046/01.py) to standardize imports across the repo.
| class Solution: | ||
| def minInversionCount(self, nums: List[int], k: int) -> int: | ||
| # Compress numbers to range [1, n] | ||
| sorted_nums = sorted(set(nums)) | ||
| comp_map = {val: idx + 1 for idx, val in enumerate(sorted_nums)} | ||
| compressed = [comp_map[num] for num in nums] | ||
| n = len(compressed) | ||
|
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||
| # Fenwick Tree (Binary Indexed Tree) for counting | ||
| class FenwickTree: | ||
| def __init__(self, size): | ||
| self.n = size | ||
| self.tree = [0] * (size + 1) | ||
|
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||
| def update(self, idx, delta): | ||
| while idx <= self.n: | ||
| self.tree[idx] += delta | ||
| idx += idx & -idx | ||
|
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||
| def query(self, idx): | ||
| res = 0 | ||
| while idx > 0: | ||
| res += self.tree[idx] | ||
| idx -= idx & -idx | ||
| return res | ||
|
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||
| # Initialize Fenwick Tree | ||
| ft = FenwickTree(n) | ||
|
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||
| # Calculate inversion count for first window | ||
| inv_count = 0 | ||
| for i in range(k): | ||
| # Count numbers greater than compressed[i] in current window | ||
| inv_count += i - ft.query(compressed[i]) | ||
| ft.update(compressed[i], 1) | ||
|
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||
| res = inv_count | ||
|
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| # Slide window | ||
| for i in range(k, n): | ||
| # Remove leftmost element | ||
| ft.update(compressed[i - k], -1) | ||
| # Count numbers smaller than removed element | ||
| removed = compressed[i - k] | ||
| smaller = ft.query(removed - 1) | ||
| inv_count -= smaller | ||
|
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||
| # Add new element | ||
| new_val = compressed[i] | ||
| # Count numbers greater than new element in current window | ||
| greater = (k - 1) - ft.query(new_val) | ||
| inv_count += greater | ||
| ft.update(new_val, 1) | ||
|
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||
| res = min(res, inv_count) | ||
|
|
||
| return res |
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Add missing List import from typing
List is used in the type annotation on line 2 but is never imported. This will raise a NameError at runtime. Add the import at the top of the file:
+from typing import List
+
class Solution:
def minInversionCount(self, nums: List[int], k: int) -> int:The algorithm itself—coordinate compression, Fenwick Tree implementation, first-window inversion count initialization, and sliding-window updates—is correct and efficiently computes the minimum inversion count across all subarrays of size k.
📝 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.
| class Solution: | |
| def minInversionCount(self, nums: List[int], k: int) -> int: | |
| # Compress numbers to range [1, n] | |
| sorted_nums = sorted(set(nums)) | |
| comp_map = {val: idx + 1 for idx, val in enumerate(sorted_nums)} | |
| compressed = [comp_map[num] for num in nums] | |
| n = len(compressed) | |
| # Fenwick Tree (Binary Indexed Tree) for counting | |
| class FenwickTree: | |
| def __init__(self, size): | |
| self.n = size | |
| self.tree = [0] * (size + 1) | |
| def update(self, idx, delta): | |
| while idx <= self.n: | |
| self.tree[idx] += delta | |
| idx += idx & -idx | |
| def query(self, idx): | |
| res = 0 | |
| while idx > 0: | |
| res += self.tree[idx] | |
| idx -= idx & -idx | |
| return res | |
| # Initialize Fenwick Tree | |
| ft = FenwickTree(n) | |
| # Calculate inversion count for first window | |
| inv_count = 0 | |
| for i in range(k): | |
| # Count numbers greater than compressed[i] in current window | |
| inv_count += i - ft.query(compressed[i]) | |
| ft.update(compressed[i], 1) | |
| res = inv_count | |
| # Slide window | |
| for i in range(k, n): | |
| # Remove leftmost element | |
| ft.update(compressed[i - k], -1) | |
| # Count numbers smaller than removed element | |
| removed = compressed[i - k] | |
| smaller = ft.query(removed - 1) | |
| inv_count -= smaller | |
| # Add new element | |
| new_val = compressed[i] | |
| # Count numbers greater than new element in current window | |
| greater = (k - 1) - ft.query(new_val) | |
| inv_count += greater | |
| ft.update(new_val, 1) | |
| res = min(res, inv_count) | |
| return res | |
| from typing import List | |
| class Solution: | |
| def minInversionCount(self, nums: List[int], k: int) -> int: | |
| # Compress numbers to range [1, n] | |
| sorted_nums = sorted(set(nums)) | |
| comp_map = {val: idx + 1 for idx, val in enumerate(sorted_nums)} | |
| compressed = [comp_map[num] for num in nums] | |
| n = len(compressed) | |
| # Fenwick Tree (Binary Indexed Tree) for counting | |
| class FenwickTree: | |
| def __init__(self, size): | |
| self.n = size | |
| self.tree = [0] * (size + 1) | |
| def update(self, idx, delta): | |
| while idx <= self.n: | |
| self.tree[idx] += delta | |
| idx += idx & -idx | |
| def query(self, idx): | |
| res = 0 | |
| while idx > 0: | |
| res += self.tree[idx] | |
| idx -= idx & -idx | |
| return res | |
| # Initialize Fenwick Tree | |
| ft = FenwickTree(n) | |
| # Calculate inversion count for first window | |
| inv_count = 0 | |
| for i in range(k): | |
| # Count numbers greater than compressed[i] in current window | |
| inv_count += i - ft.query(compressed[i]) | |
| ft.update(compressed[i], 1) | |
| res = inv_count | |
| # Slide window | |
| for i in range(k, n): | |
| # Remove leftmost element | |
| ft.update(compressed[i - k], -1) | |
| # Count numbers smaller than removed element | |
| removed = compressed[i - k] | |
| smaller = ft.query(removed - 1) | |
| inv_count -= smaller | |
| # Add new element | |
| new_val = compressed[i] | |
| # Count numbers greater than new element in current window | |
| greater = (k - 1) - ft.query(new_val) | |
| inv_count += greater | |
| ft.update(new_val, 1) | |
| res = min(res, inv_count) | |
| return res |
🧰 Tools
🪛 Ruff (0.14.7)
2-2: Undefined name List
(F821)
🤖 Prompt for AI Agents
In solutions/3768/01.py around lines 1 to 57, the type annotation uses List but
the symbol is not imported; add "from typing import List" at the top of the file
(as the first import) so the type annotation resolves, then run a quick
lint/type check to confirm no other missing imports.
| class Solution: | ||
| def totalScore(self, hp: int, damage: List[int], requirement: List[int]) -> int: | ||
| n = len(damage) | ||
|
|
||
| # Prefix sums for damage | ||
| prefix = [0] * (n + 1) | ||
| for i in range(n): | ||
| prefix[i + 1] = prefix[i] + damage[i] | ||
|
|
||
| # For each starting position i, calculate score | ||
| res = 0 | ||
| for i in range(n): | ||
| current_hp = hp | ||
| score = 0 | ||
| for j in range(i, n): | ||
| current_hp -= damage[j] | ||
| if current_hp >= requirement[j]: | ||
| score += 1 | ||
| res += score | ||
|
|
||
| return res |
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Add missing List import and remove unused prefix sums code.
-
Listin the signature (line 2) is undefined; add the required import:+from typing import List + class Solution:
-
The
prefixarray (lines 6–8) is unused dead code that should be removed.
The O(n²) complexity is unavoidable for this problem given the constraints (n ≤ 10⁵): you must check each (i, j) pair where i is the starting room and j ranges from i to n−1. Prefix sums do not improve the worst-case complexity.
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| class Solution: | |
| def totalScore(self, hp: int, damage: List[int], requirement: List[int]) -> int: | |
| n = len(damage) | |
| # Prefix sums for damage | |
| prefix = [0] * (n + 1) | |
| for i in range(n): | |
| prefix[i + 1] = prefix[i] + damage[i] | |
| # For each starting position i, calculate score | |
| res = 0 | |
| for i in range(n): | |
| current_hp = hp | |
| score = 0 | |
| for j in range(i, n): | |
| current_hp -= damage[j] | |
| if current_hp >= requirement[j]: | |
| score += 1 | |
| res += score | |
| return res | |
| from typing import List | |
| class Solution: | |
| def totalScore(self, hp: int, damage: List[int], requirement: List[int]) -> int: | |
| n = len(damage) | |
| # For each starting position i, calculate score | |
| res = 0 | |
| for i in range(n): | |
| current_hp = hp | |
| score = 0 | |
| for j in range(i, n): | |
| current_hp -= damage[j] | |
| if current_hp >= requirement[j]: | |
| score += 1 | |
| res += score | |
| return res |
🧰 Tools
🪛 Ruff (0.14.7)
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
🤖 Prompt for AI Agents
In solutions/3771/01.py around lines 1 to 21, the function signature uses List
but the typing.List import is missing and the prefix sums array is unused dead
code; fix by adding "from typing import List" at the top of the file (or include
List in existing typing imports) and remove the prefix array creation and its
population (lines that create and fill prefix) so only the necessary O(n²) loops
remain.
| class Solution: | ||
| def maxSubgraphScore(self, n: int, edges: List[List[int]], good: List[int]) -> List[int]: | ||
| # Build tree | ||
| from collections import defaultdict | ||
| graph = defaultdict(list) | ||
| for u, v in edges: | ||
| graph[u].append(v) | ||
| graph[v].append(u) | ||
|
|
||
| # Rerooting DP | ||
| # dp[node] = max score of connected subgraph containing node | ||
| # We use tree DP with rerooting | ||
|
|
||
| # First, calculate for each node the best subtree score | ||
| def dfs(node, parent): | ||
| # Score contribution of this node | ||
| node_score = 1 if good[node] else -1 | ||
|
|
||
| # Best score including this node and its children | ||
| best = node_score | ||
| for child in graph[node]: | ||
| if child != parent: | ||
| child_score = dfs(child, node) | ||
| # Include child if it improves the score | ||
| if child_score > 0: | ||
| best += child_score | ||
| return best | ||
|
|
||
| # For each node as root, calculate max score | ||
| res = [] | ||
| for root in range(n): | ||
| score = dfs(root, -1) | ||
| res.append(score) | ||
|
|
||
| return res |
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🧩 Analysis chain
🏁 Script executed:
cat -n solutions/3772/01.pyRepository: romankurnovskii/leetcode-apps
Length of output: 1555
🏁 Script executed:
# Look for explanation or constraints
find solutions/3772 -type f | head -20Repository: romankurnovskii/leetcode-apps
Length of output: 93
🌐 Web query:
LeetCode 3772 maxSubgraphScore constraints
💡 Result:
I couldn't find a LeetCode problem titled "maxSubgraphScore" or number 3772 with that name. Do you mean a different problem title or number (e.g., "maximum subgraph score", a contest problem, or a problem ID)? If you want, I can search LeetCode and other sources again — tell me the exact problem link, title, or paste the statement.
Add missing typing import and address O(n²) complexity despite "rerooting DP" claim.
Listin the signature (line 2) is undefined; add the import:
+from typing import List
+
class Solution:
def maxSubgraphScore(self, n: int, edges: List[List[int]], good: List[int]) -> List[int]:-
Despite the "Rerooting DP" comment (lines 10–12), the code runs a full
dfsfor every root (lines 31–33), yielding O(n²) time complexity. True rerooting DP computes answers in O(n) or O(n log n) via two passes: one post-order DFS to compute subtree scores, then one pre-order DFS to propagate changes across roots. This implementation does not match that pattern. -
The recursive
dfscan exceed Python's recursion limit (~1000) on deep trees if n is large.
Consider either implementing true rerooting DP to match the explanation and achieve O(n) complexity, or explicitly document that this is a brute-force O(n²) solution.
🧰 Tools
🪛 Ruff (0.14.7)
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/3772/01.py around lines 1–35, fix three issues: add the missing
typing import (e.g., "from typing import List") at the top; replace the current
O(n²) approach (calling dfs for every root) with a true rerooting DP: perform
one post-order DFS to compute dp_down[node] = max score of connected subgraph
inside node's subtree including node, then a pre-order DFS to propagate
contributions and compute final_answer[node] by re-rooting using parent-to-child
transfers (avoid re-running full DFS per root); and handle deep recursion by
either setting sys.setrecursionlimit to a safe value or converting the DFSs to
iterative stacks. Ensure the final result uses the two-pass rerooting values and
remove the per-root dfs loop.
2 participants
Summary
This PR adds solutions and explanations for 15 LeetCode problems:
Problems Added
2390 - Removing Stars From a String (Medium, Stack)
2542 - Maximum Subsequence Score (Medium, Heap)
3753 - Paths in Matrix Whose Sum Is Divisible by K (Hard, DP)
3759 - Minimum Cost to Convert String I (Medium, Graph)
3760 - Minimum Cost to Convert String II (Hard, Graph/DP)
3761 - Count Valid Paths in a Tree (Hard, Tree DP)
3762 - Minimum Operations to Make Array Values Equal to K (Easy, Array)
3765 - Complete Prime Number (Medium, Math)
3766 - Minimum Operations to Make Binary Palindrome (Medium, Array)
3767 - Maximize Points After Choosing K Tasks (Medium, Greedy)
3768 - Minimum Inversion Count in Subarrays of Fixed Length (Hard, Fenwick Tree)
3769 - Sort Integers by Binary Reflection (Easy, Sorting)
3770 - Largest Prime from Consecutive Prime Sum (Medium, Math)
3771 - Total Score of Dungeon Runs (Medium, Simulation)
3772 - Maximum Subgraph Score in a Tree (Hard, Tree DP)
Approach
All solutions follow the standard workflow:
resas the result variableNote
Solutions have been created and explanations written. Submissions to LeetCode may need to be verified separately due to session expiration during initial testing.
Summary by Sourcery
Add implementations and written explanations for several LeetCode problems, including stack, greedy/heap, Fenwick tree, tree DP, and math-based solutions, and align an existing solution with the standardized explanation/result-variable pattern.
New Features:
Enhancements:
Summary by CodeRabbit
New Features
Documentation
✏️ Tip: You can customize this high-level summary in your review settings.