#757

Hard

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Set Intersection Size At Least Two

Solve the Set Intersection Size At Least Two problem using a greedy approach and invariant validation.

Problem Statement

Given an array of intervals, where each interval consists of two integers representing the inclusive range of numbers, your goal is to find a set of integers that intersects each interval in the array with at least two elements.

Return the minimum size of such a set that meets the condition for all intervals. The problem emphasizes using greedy choices along with invariant validation to minimize the set size while satisfying the requirements of the intervals.

Examples

Example 1

Input: intervals = [[1,3],[3,7],[8,9]]

Output: 5

let nums = [2, 3, 4, 8, 9]. It can be shown that there cannot be any containing array of size 4.

Example 2

Input: intervals = [[1,3],[1,4],[2,5],[3,5]]

Output: 3

let nums = [2, 3, 4]. It can be shown that there cannot be any containing array of size 2.

Example 3

Input: intervals = [[1,2],[2,3],[2,4],[4,5]]

Output: 5

let nums = [1, 2, 3, 4, 5]. It can be shown that there cannot be any containing array of size 4.

Constraints

1 <= intervals.length <= 3000
intervals[i].length == 2
0 <= starti < endi <= 108

Solution Approach

Sort and Greedy Selection

Start by sorting the intervals based on their end points. Then, iteratively choose elements from the sorted intervals that satisfy the condition of at least two elements intersecting with each interval.

Invariant Validation

For each interval, ensure the selected elements maintain the invariant that at least two points are covered. This is validated by checking the current selected elements against the interval ranges and adjusting the set when necessary.

Optimal Set Construction

Construct the optimal set of integers by choosing points that cover the most number of intervals. After each selection, validate that the set still contains at least two elements for each interval.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the sorting step, which is O(n log n), where n is the number of intervals. The space complexity is O(n), as we store the selected integers in the set.

What Interviewers Usually Probe

The candidate demonstrates an understanding of greedy strategies and interval problems.
The candidate considers both the sorting step and the invariant maintenance when explaining the solution.
The candidate emphasizes validating the greedy choice and its impact on minimizing the set size.

Common Pitfalls or Variants

Common pitfalls

Choosing too few elements without validating the intersection for all intervals.
Failing to maintain the invariant that at least two elements must intersect each interval.
Not sorting the intervals properly, leading to inefficient greedy selections.

Follow-up variants

What if we had to make the set size exactly two for each interval?
How would the solution change if intervals could overlap in more complex ways?
What is the effect of a larger number of intervals on the overall set size?

FAQ

What is the greedy approach for Set Intersection Size At Least Two?

The greedy approach involves selecting key elements from sorted intervals to ensure that at least two elements are covered for each interval, minimizing the set size.

How do I validate the set intersection for each interval?

After selecting the elements for the set, check if every interval contains at least two elements from the set. If not, adjust the set accordingly.

What is the time complexity of the Set Intersection Size At Least Two problem?

The time complexity is dominated by the sorting step, which is O(n log n), where n is the number of intervals.

Can this problem be solved without sorting the intervals?

Sorting the intervals is crucial for ensuring the greedy approach works efficiently. Without sorting, the problem would likely require more complex logic and potentially slower solutions.

What if the problem requires more than two elements in the intersection?

The solution would need to adjust the greedy strategy to select the appropriate number of elements to satisfy the new condition, but the core approach would remain similar.

terminal

Solution

Solution 1: Sorting + Greedy

We want to select as few integer points as possible on the number line such that each interval contains at least two points. A classic and effective strategy is to sort intervals by their right endpoints and try to place selected points towards the right side of intervals, so that these points can cover more subsequent intervals.

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class Solution:
    def intersectionSizeTwo(self, intervals: List[List[int]]) -> int:
        intervals.sort(key=lambda x: (x[1], -x[0]))
        s = e = -1
        ans = 0
        for a, b in intervals:
            if a <= s:
                continue
            if a > e:
                ans += 2
                s, e = b - 1, b
            else:
                ans += 1
                s, e = e, b
        return ans

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