Asked in: DPWORLD
Image of the Question
All Testcases Passed ✔
Solution
import bisect
def interval(first, last):
arr = []
for i,j in zip(first, last):
// ... rest of solution available after purchaseExplanation
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To approach the problem of finding the smallest set of integers such that every interval contains at least two integers from that set, it’s important to understand the problem's essence and what it demands. We are given a list of intervals defined by their start and end points, and our goal is to select the minimal number of integers to cover these intervals so that each interval has at least two integers inside it.
Step 1: Understand the Requirements
Each interval must have at least two integers from the chosen set inside it. For example, if an interval is [3, 5], it means the chosen integers must include any two integers from 3, 4, and 5. The key is to minimize the total number of integers chosen over all intervals while satisfying this condition for every interval.
Step 2: Analyze the Nature of the Problem
This is a type of coverage problem similar to "set cover" but with an added constraint that each interval requires two points instead of one. Simply picking one integer per interval won’t work here; instead, each interval imposes a minimum coverage of two integers. The intervals may overlap, so cleverly choosing integers that cover multiple intervals simultaneously is the key to minimizing the total count.
Step 3: Consider Sorting the Intervals
To manage overlapping intervals and efficiently cover them, sort the intervals by their end points in ascending order. Sorting by end helps because it guides a greedy approach to picking points towards the right end of the intervals, which may cover subsequent intervals more effectively. This sorting aligns with classic interval covering strategies where working from earliest finishing intervals to later ones helps in minimizing the coverage set.
Step 4: Greedy Strategy to Pick Points
After sorting intervals by their end values, start processing each interval one by one. For each interval, check how many points from your chosen set currently lie within it.
- If the interval already has two or more points from the chosen set, no additional points need to be added.
- If there is only one point inside, add one more point to meet the two-point requirement.
- If no points are inside, add two points to cover it.
To decide which points to add, pick the largest possible points near the interval’s end to maximize overlap with future intervals. For example, if an interval ends at 'end', choose points at 'end-1' and 'end' if both fit inside the interval. This choice ensures those points may also cover the next intervals that might start before or at 'end', reducing the total number of added points.
Step 5: Efficiently Track Chosen Points
Keep track of the points added so far and efficiently check their membership in upcoming intervals. Since the problem can involve up to 10^5 intervals, efficiency is crucial. Maintain a structure or variables to hold the last two chosen points as you move through intervals because only the most recent chosen points are relevant for checking coverage in the sorted order.
Step 6: Example Walkthrough
Consider intervals: [0, 2], [1, 3], and [2, 3]. Sorted by end:
- [0, 2] ends at 2
- [1, 3] ends at 3
- [2, 3] ends at 3
Process [0, 2]: No points chosen yet, so add two points: 1 and 2.
Process [1, 3]: Points chosen so far are 1 and 2. Both fall inside [1, 3], so no new points needed.
Process [2, 3]: Points chosen are still 1 and 2. Only 2 lies inside [2, 3], so one more point must be added, choose 3. Now the chosen set is {1, 2, 3}.
Step 7: Complexity Considerations
The algorithm should process each interval once after sorting, making it O(n log n) because of the sorting step and O(n) for the main logic. Checking coverage against the last chosen points is O(1) per interval. Avoiding extra complex data structures helps to meet performance requirements.
Step 8: Edge Cases to Consider
- Intervals that are adjacent or overlapping heavily.
- Intervals with very large start and end values to ensure point selection near ends is consistent.
- Intervals with minimum length (e.g., difference of 1) where only two integers fit exactly.
- Large number of intervals requiring optimal coverage to avoid choosing excessive points.
Step 9: Summary
1. Sort intervals by their end points to process them from earliest finish to latest.
2. For each interval, count how many chosen points already lie inside it.
3. If less than two points cover the interval, add the minimal number of points near the interval’s end to make it two.
4. Keep track of the last two points added to optimize checks for subsequent intervals.
5. Continue this process through all intervals.
6. The final number of points chosen is the answer — the size of the smallest set meeting the requirement.
By focusing on a greedy strategy that selects points near interval ends after sorting, you ensure maximum overlap of coverage between intervals, reducing the total points needed and providing an efficient and correct solution.
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