3439. Reschedule Meetings for Maximum Free Time I #253
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Reviewer's GuideAdds sliding window solutions for substring rearrangement problems and a gap-based sliding window algorithm to compute maximum free time across k consecutive gaps in meeting schedules. Class diagram for new Solution classesclassDiagram
class Solution_3297 {
+long long validSubstringCount(string word1, string word2)
}
class Solution_3298 {
+long long validSubstringCount(string word1, string word2)
}
class Solution_3439 {
+int maxFreeTime(int eventTime, int k, vector<int>& startTime, vector<int>& endTime)
}
Flow diagram for gap-based sliding window in maxFreeTimeflowchart TD
A["Input: eventTime, k, startTime[], endTime[]"] --> B["Compute gaps between events (temp[])"]
B --> C["Initialize sliding window pointers l, r"]
C --> D["Iterate r over temp[]"]
D --> E["Add temp[r] to tempsum"]
E --> F["If window size == k"]
F --> G["Update maxsum if tempsum > maxsum"]
G --> H["Subtract temp[l] from tempsum and increment l"]
H --> D
D -->|r reaches end| I["Return maxsum"]
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Hey there - I've reviewed your changes - here's some feedback:
- This PR combines three unrelated problem solutions—please split it into separate PRs so each submission remains focused and easier to review.
- In maxFreeTime(), the sliding‐window check
if ((r - l) == k)currently sums k+1 gaps; double‐check the window boundaries so you only include exactly k gaps per window. - The two validSubstringCount implementations are identical duplicates; consider extracting the logic into a shared helper or common function to reduce repetition.
Prompt for AI Agents
Please address the comments from this code review:
## Overall Comments
- This PR combines three unrelated problem solutions—please split it into separate PRs so each submission remains focused and easier to review.
- In maxFreeTime(), the sliding‐window check `if ((r - l) == k)` currently sums k+1 gaps; double‐check the window boundaries so you only include exactly k gaps per window.
- The two validSubstringCount implementations are identical duplicates; consider extracting the logic into a shared helper or common function to reduce repetition.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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Intuition
We are asked to find the maximum total free time across k consecutive gaps between events.
The key idea is to first compute all gaps between consecutive events (including the gap before the first event and after the last event). Once we have these gaps, the problem reduces to finding the maximum sum of any k consecutive gaps, which can be efficiently done using a sliding window.
Approach
Compute gaps:
The first gap is the time from 0 to the start of the first event.
The last gap is the time from the end of the last event to eventTime.
For all intermediate gaps, calculate startTime[i] - endTime[i-1].
Sliding window to find maximum sum:
Use two pointers l (left) and r (right) to maintain a window of size k.
Keep a running sum of the gaps in the current window.
As the window moves forward, update the sum by adding the new gap and removing the leftmost gap.
Track the maximum sum seen so far.
Return the result:
After processing all windows, the maximum sum represents the largest total free time for k consecutive gaps.
Code Solution (C++)
class Solution {
public:
int maxFreeTime(int eventTime, int k, vector& startTime, vector& endTime) {
int n = startTime.size();
vector temp(n + 1, 0);
temp[0] = startTime[0] - 0;
temp[n] = eventTime - endTime[n - 1];
for (int i = 1; i < n; i++) {
temp[i] = startTime[i] - endTime[i - 1];
}
int l = 0, r = 0, tempsum = 0, maxsum = 0;
while (r < temp.size()) {
tempsum += temp[r];
if ((r - l) == k) {
maxsum = max(maxsum, tempsum);
tempsum -= temp[l];
l++;
}
r++;
}
return maxsum;
}
};
Related Issues
Closes #252
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Summary by Sourcery
Add three new C++ solutions for LeetCode problems: two substring rearrangement count variants and a meeting rescheduling free-time maximization using sliding window methods.
New Features: