Add solutions and explanations for problems 3652, 3769, 3774, 3775, 3776, 3777 #123
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…ell Stock using Strategy
Reviewer's GuideAdds solutions and structured explanations for LeetCode problems 3652, 3769, 3774, 3775, 3776, and 3777, while refining several existing solutions and explanations for clarity, correctness, and implementation detail (notably for problems 3769, 3774, 3775, 3776, and 3777). Class diagram for newly added and updated Solution classes and FenwickTreeclassDiagram
class Solution3652 {
+maxProfit(prices, strategy, k) int
}
class Solution3769 {
+sortByReflection(nums) List~int~
+binary_reflection(n) int
}
class Solution3774 {
+absDifference(nums, k) int
}
class Solution3775 {
+reverseWords(s) str
+count_vowels(word) int
}
class Solution3776 {
+minMoves(balance) int
}
class Solution3777 {
+minDeletions(s, queries) List~int~
}
class FenwickTree {
-tree List~int~
+FenwickTree(size)
+add(i, delta) void
+query(i) int
}
Solution3777 --> FenwickTree : uses
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WalkthroughThis PR adds a new explanation for problem 3652 and updates explanations and solutions for problems 3652, 3769, 3774, 3775, 3776, and 3777. Changes primarily consist of documentation rewording, clarification, and structural improvements to explanations, along with minor solution refinements including variable renaming, logic adjustments, and comment enhancements. Changes
Estimated code review effort🎯 3 (Moderate) | ⏱️ ~20 minutes
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Hey there - I've reviewed your changes - here's some feedback:
- In
solutions/3777/01.py, the Fenwick tree updates on flips (bit.add(i, 1 if A[i] == A[i - 1] else -1)etc.) don't account for the previous violation state at those positions, so repeated flips will accumulate incorrect counts; consider computing the old and new violation flags and updating the tree by their difference instead. - The implementation in
solutions/3776/01.pyonly considers the first negative index and then unconditionally returnsreswithout checking if all negative balances are fully covered, which can miss cases with multiple negative positions or insufficient nearby positives; consider generalizing to handle all negative balances and reintroducing an impossibility check when remaining demand is nonzero. - For problem 3652, the added explanation describes a binary-string flipping problem while
solutions/3652/01.pyimplements a stock strategy/max profit function, so the problem statement and solution appear mismatched and should be aligned to the same task.
Prompt for AI Agents
Please address the comments from this code review:
## Overall Comments
- In `solutions/3777/01.py`, the Fenwick tree updates on flips (`bit.add(i, 1 if A[i] == A[i - 1] else -1)` etc.) don't account for the previous violation state at those positions, so repeated flips will accumulate incorrect counts; consider computing the old and new violation flags and updating the tree by their difference instead.
- The implementation in `solutions/3776/01.py` only considers the first negative index and then unconditionally returns `res` without checking if all negative balances are fully covered, which can miss cases with multiple negative positions or insufficient nearby positives; consider generalizing to handle all negative balances and reintroducing an impossibility check when remaining demand is nonzero.
- For problem 3652, the added explanation describes a binary-string flipping problem while `solutions/3652/01.py` implements a stock strategy/max profit function, so the problem statement and solution appear mismatched and should be aligned to the same task.Help me be more useful! Please click 👍 or 👎 on each comment and I'll use the feedback to improve your reviews.
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Actionable comments posted: 2
🧹 Nitpick comments (1)
solutions/3776/01.py (1)
39-44: Consider unpacking only needed variables.The loop variable
idxis extracted but never used within the loop body. You can simplify the unpacking.🔎 Apply this diff to unpack only needed variables:
- for dist, idx, amount in positives: + for dist, _idx, amount in positives: if remaining <= 0: break transfer = min(remaining, amount)Alternatively, if the index is truly not needed for debugging or future use, you can unpack with:
for dist, _, amount in positives:
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Review profile: CHILL
Plan: Pro
📒 Files selected for processing (12)
explanations/3652/en.md(1 hunks)explanations/3769/en.md(1 hunks)explanations/3774/en.md(1 hunks)explanations/3775/en.md(1 hunks)explanations/3776/en.md(1 hunks)explanations/3777/en.md(1 hunks)solutions/3652/01.py(1 hunks)solutions/3769/01.py(1 hunks)solutions/3774/01.py(1 hunks)solutions/3775/01.py(1 hunks)solutions/3776/01.py(1 hunks)solutions/3777/01.py(2 hunks)
🧰 Additional context used
🪛 LanguageTool
explanations/3775/en.md
[grammar] ~24-~24: Use a hyphen to join words.
Context: ...s in a single pass and reusing the vowel counting logic, we avoid redundant opera...
(QB_NEW_EN_HYPHEN)
🪛 Ruff (0.14.8)
solutions/3776/01.py
39-39: Loop control variable idx not used within loop body
Rename unused idx to _idx
(B007)
solutions/3652/01.py
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
⏰ 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)
- GitHub Check: Sourcery review
🔇 Additional comments (11)
solutions/3769/01.py (1)
3-11: LGTM! Readability improvements.The added comments and explicit result variable improve code clarity without altering functionality.
solutions/3774/01.py (1)
2-15: LGTM! Clear implementation with good comments.The refactoring to use
sorted()and explicit variable names improves readability and avoids mutating the input array.solutions/3777/01.py (1)
1-49: LGTM! Enhanced documentation.The added comments clarify the Fenwick Tree operations and violation tracking logic without changing behavior.
solutions/3775/01.py (1)
3-25: LGTM! Clean refactoring with improved edge case handling.The helper function improves code organization, and returning
swhen there are no words (line 9) correctly handles edge cases like empty strings or strings with only whitespace.solutions/3776/01.py (1)
1-46: Distance-based greedy approach looks correct.The refactored circular distance calculation and nearest-positive transfer logic appear sound for balancing the circular array.
explanations/3775/en.md (1)
1-61: LGTM! Clear and well-structured explanation.The documentation effectively explains the problem approach with good examples and complexity analysis.
solutions/3652/01.py (1)
3-34: Prefix sum optimization looks correct.The algorithm efficiently computes the maximum profit by:
- Calculating base profit without modifications
- Using prefix sums to evaluate each segment in O(1)
- Finding the best segment to modify
The logic for computing
modified_segment_profit(treating first k/2 as hold, last k/2 as sell) aligns with the strategy transformation approach.explanations/3769/en.md (1)
1-61: Clear and pedagogically sound explanation of binary reflection sorting.The algorithm, example walkthrough, and complexity analysis are all accurate. The composite sort key (binary_reflection, original_value) is well-motivated, and the trace table clearly demonstrates how the sorting works. The edge case handling for tied reflections is correctly noted.
explanations/3774/en.md (1)
1-60: Accurate and clear explanation of the sorting-based sum difference approach.The algorithm is correctly explained, the example walkthrough is mathematically correct ([2,2,4,5] → sums 4 and 9 → difference 5), and the complexity analysis is sound. The edge case note about k=n is appropriate.
explanations/3777/en.md (1)
1-64: Well-structured Fenwick Tree explanation with accurate violation-tracking model.The algorithm correctly models violations as adjacent identical character pairs, and using a Fenwick Tree for range queries is a sound optimization. The example walkthrough accurately traces the state: "ABA" initially has 0 violations, flipping index 1 to 'A' creates "AAA" with 2 violations (pairs [0,1] and [1,2]), correctly resulting in 2 deletions needed. The complexity analysis O(n + q log n) is accurate.
explanations/3776/en.md (1)
1-65: Correct greedy strategy with accurate distance calculations.The circular distance formula min(|i - neg_idx|, n - |i - neg_idx|) is correct, and the greedy approach of prioritizing nearest positive balances is sound for minimizing total moves. The example correctly shows both indices 0 and 1 at distance 1 from the negative position, and the final cost of 4 moves (4 units × distance 1) is accurate.
| ## Explanation | ||
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| ### Strategy (The "Why") | ||
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| **Restate the problem:** We are given a binary string and can perform two types of operations: flip all characters from index 0 to i (cost: i+1) or flip all characters from index i to n-1 (cost: n-i). We need to find the minimum cost to make all characters equal. | ||
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| **1.1 Constraints & Complexity:** | ||
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| - **Input Size:** The string length n can be up to 10^5, and each character is either '0' or '1'. | ||
| - **Time Complexity:** O(n) - we iterate through the string once to find all transitions. | ||
| - **Space Complexity:** O(1) - we only use a constant amount of extra space. | ||
| - **Edge Case:** If the string already has all characters equal, there are no transitions and the cost is 0. | ||
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| **1.2 High-level approach:** | ||
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| The goal is to find the minimum cost to eliminate all transitions (places where adjacent characters differ) in the string. When we encounter a transition, we can either flip the prefix or the suffix, and we choose the cheaper option. | ||
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| **1.3 Brute force vs. optimized strategy:** | ||
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| - **Brute Force:** Try all possible combinations of operations, which would be exponential O(2^n) and too slow. | ||
| - **Optimized Strategy:** For each transition at position i, we can either flip [0, i-1] (cost i) or flip [i, n-1] (cost n-i). We choose min(i, n-i) and sum all transition costs. This is O(n) time. | ||
| - **Optimization:** By recognizing that each transition can be fixed independently with the cheaper operation, we avoid exponential search and solve in linear time. | ||
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| **1.4 Decomposition:** | ||
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| 1. Iterate through the string from left to right. | ||
| 2. For each position i, check if there's a transition (s[i] != s[i-1]). | ||
| 3. If a transition exists, add the minimum cost (min(i, n-i)) to fix it. | ||
| 4. Return the total cost. | ||
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| ### Steps (The "How") | ||
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| **2.1 Initialization & Example Setup:** | ||
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| Let's use the example: `s = "0011"`, `n = 4`. | ||
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| - Initialize `res = 0` to track total cost. | ||
| - The string has one transition at position 2 (between '0' and '1'). | ||
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| **2.2 Start Checking:** | ||
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| We iterate through the string starting from index 1, comparing each character with the previous one. | ||
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| **2.3 Trace Walkthrough:** | ||
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| | Position i | s[i] | s[i-1] | Transition? | Cost Option 1 (flip prefix) | Cost Option 2 (flip suffix) | Chosen Cost | res | | ||
| |------------|------|--------|-------------|---------------------------|----------------------------|-------------|-----| | ||
| | 1 | '0' | '0' | No | - | - | 0 | 0 | | ||
| | 2 | '1' | '0' | Yes | 2 | 2 | 2 | 2 | | ||
| | 3 | '1' | '1' | No | - | - | 0 | 2 | | ||
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| **2.4 Increment and Loop:** | ||
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| After checking all positions, we have found all transitions and calculated the minimum cost to fix each one. | ||
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| **2.5 Return Result:** | ||
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| The result is 2, which is the minimum cost to make all characters equal. We achieve this by flipping the suffix from index 2 to the end, changing "0011" to "0000". | ||
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🧩 Analysis chain
🏁 Script executed:
# Find the solution file for problem 3652
find . -path "*/3652/*" -type f | head -20Repository: romankurnovskii/leetcode-apps
Length of output: 121
🏁 Script executed:
# Read the solution file to see what problem it solves
if [ -f "solutions/3652/01.py" ]; then
cat -n solutions/3652/01.py
fiRepository: romankurnovskii/leetcode-apps
Length of output: 1748
🌐 Web query:
LeetCode problem 3652
💡 Result:
LeetCode 3652 — "Best Time to Buy and Sell Stock using Strategy"
Problem (brief):
- Given arrays prices and strategy (strategy[i] ∈ {-1,0,1}), and even k. You may change exactly one length-k consecutive window: set its first k/2 entries to 0 and last k/2 to 1. Profit = sum(strategy[i] * prices[i]). Return max possible profit after at most one such modification. There is no ownership/budget constraint. [1]
Key idea / O(n) solution:
- Precompute prefix sums s[i] = sum_{0..i-1} strategy[j]*prices[j] and t[i] = sum_{0..i-1} prices[j].
- Base profit = s[n].
- For each window ending at i (window [i-k, i-1]), the profit change if modified equals:
Δ = - (s[i] - s[i-k]) + (t[i] - t[i-k/2]). - Maximize s[n] + Δ over all valid i. Time O(n), space O(n). [1]
Complexity:
- Time: O(n)
- Space: O(n) (can be reduced to O(1) extra if careful). [1]
Reference:
- Problem statement and solution outline. [1]
[1] LeetCode 3652 problem page / solution summary.
Explanation file contains wrong problem description.
The explanation file explanations/3652/en.md describes a binary string flipping problem ("flip all characters from index 0 to i"), but LeetCode problem 3652 is "Best Time to Buy and Sell Stock using Strategy"—a stock trading profit maximization problem. The solution file solutions/3652/01.py correctly implements the stock trading problem with maxProfit(prices, strategy, k).
Replace the explanation with content matching the actual problem: given arrays prices, strategy (with values -1, 0, 1), and even k, modify exactly one length-k consecutive window (first k/2 entries to 0, last k/2 to 1) to maximize profit where profit = sum(strategy[i] * prices[i]).
🤖 Prompt for AI Agents
In explanations/3652/en.md around lines 1 to 61, the current content describes a
binary string flipping problem but must describe LeetCode 3652 (Best Time to Buy
and Sell Stock using Strategy); replace the entire file with an explanation for
the stock problem: state the problem (inputs: prices array, strategy array with
values -1/0/1, integer k and the allowed operation of selecting one length-k
window and setting first floor(k/2) entries of that window to 0 and last
ceil(k/2) to 1), include constraints (n up to 1e5, values ranges), desired time
O(n) and space O(1), present the optimized approach (compute base profit
sum(strategy[i]*prices[i]), compute the gain delta for converting each window
using sliding window of length k by summing (new_strategy - old_strategy)*prices
over window, track max gain and return base+max_gain), and include a short
worked example that matches the algorithm and the solution file behavior.
| class Solution: | ||
| def maxProfit(self, prices: List[int], strategy: List[int], k: int) -> int: |
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Missing import for type hints.
The List type hint is used but not imported, which will cause a NameError at runtime when Python evaluates the annotations.
🔎 Apply this diff to add the missing import:
+from typing import List
+
class Solution:
def maxProfit(self, prices: List[int], strategy: List[int], k: int) -> int:📝 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 maxProfit(self, prices: List[int], strategy: List[int], k: int) -> int: | |
| from typing import List | |
| class Solution: | |
| def maxProfit(self, prices: List[int], strategy: List[int], k: int) -> int: |
🧰 Tools
🪛 Ruff (0.14.8)
2-2: Undefined name List
(F821)
2-2: Undefined name List
(F821)
🤖 Prompt for AI Agents
In solutions/3652/01.py around lines 1 to 2, the function uses the List type for
annotations but List is not imported; add the missing import from typing by
adding "from typing import List" at the top of the file (before the class
definition) so the type hint resolves correctly.
romankurnovskii
deleted the
problems-3652-3705-3769-3774-3775-3776-3777
branch
2 participants
Summary
This PR adds solutions and explanations for 6 LeetCode problems:
Problem 3652: Best Time to Buy and Sell Stock using Strategy (Medium)
Problem 3769: Sort Integers by Binary Reflection (Easy)
Problem 3774: Absolute Difference Between Maximum and Minimum K Elements (Easy)
Problem 3775: Reverse Words With Same Vowel Count (Medium)
Problem 3776: Minimum Moves to Balance Circular Array (Medium)
Problem 3777: Minimum Deletions to Make Alternating Substring (Hard)
Files Added/Modified
solutions/{problem_number}/01.pyexplanations/{problem_number}/en.mdAll explanations follow the structured format defined in the workflow rules.
Summary by Sourcery
Add a new solution and explanation for problem 3652 and refine existing solutions and explanations for problems 3769, 3774, 3775, 3776, and 3777.
New Features:
Enhancements:
Summary by CodeRabbit
New Features
Documentation
Refactor
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