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Set Intersection Size At Least Two  Leetcode Solution
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LeetCode: Set Intersection Size At Least Two Leetcode Solution
Difficulty: Hard
The Set Intersection Size At Least Two problem on LeetCode can be solved using the Greedy approach. The problem can be restated as follows:
Given a set of intervals, find the minimum number of elements required to be present in every interval such that the intersection of all the intervals has a size of at least two.
To solve the problem, we can follow the following steps:

Sort the intervals based on their end points.

Assign two variables, last and secondLast, to keep track of the last two elements of the minimumsized intersection we must create. Initially, last can be set to 1 and secondLast can be set to 2.

Iterate over each interval in sorted order. Let's use the variable curr to represent the current interval.

If curr[0] > last, it means that the current interval does not overlap with the minimumsized intersection we have created so far. In this case, we need to add both curr[1]  1 and curr[1] to the minimumsized intersection, and update last and secondLast accordingly.

Otherwise, if curr[0] <= last, it means that the current interval overlaps with the minimumsized intersection. In this case, we only need to add curr[1] to the minimumsized intersection.

Our final result will be the size of the minimumsized intersection.
Here's the Python code implementing this algorithm:
def intersectionSizeTwo(intervals):
intervals.sort(key=lambda x: x[1])
last, secondLast = 1, 2
result = 0
for curr in intervals:
if curr[0] > last:
result += 2
secondLast, last = last, curr[1]  1
elif curr[0] <= secondLast:
continue
else:
result += 1
secondLast, last = last, curr[1]
return result
The time complexity of this algorithm is O(n log n), where n is the number of intervals, due to the sorting operation. The space complexity is O(1), as we only need constant space to store our variables.
Set Intersection Size At Least Two Solution Code
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