How do you solve Set Intersection Size At Least Two on LeetCode?

Set Intersection Size At Least Two (LeetCode #757) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Greedy algorithms often provide efficient solutions for covering problems.

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

The optimal solution for Set Intersection Size At Least Two runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) because we store the integers in a list.

What is the brute force solution for Set Intersection Size At Least Two?

The brute force approach for Set Intersection Size At Least Two is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible sets of integers that can cover the intervals and checking which of these sets meets the requirement of having at least two integers for each interval. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Set Intersection Size At Least Two?

Set Intersection Size At Least Two uses the following concepts: Array, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Set Intersection Size At Least Two?

Always think about sorting and greedy strategies for covering problems. Practice explaining your thought process clearly as you work through the problem. Consider edge cases and test your solution with them to ensure robustness.

What are common mistakes when solving Set Intersection Size At Least Two?

Failing to consider edge cases where intervals overlap significantly. Not properly managing the last added integer, leading to incorrect coverage.

#757

Set Intersection Size At Least Two

Hard

ArrayGreedySortingGreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal approach leverages a greedy strategy. By sorting the intervals and ensuring that we always add the largest possible integers to our set, we can minimize the size of the containing set while still satisfying the requirement of covering each interval with at least two integers.

⚙️

Algorithm

4 steps

1Step 1: Sort the intervals by their end points.
2Step 2: Initialize an empty set to hold the integers and a variable to track the last added integer.
3Step 3: Iterate through the sorted intervals and for each interval, check if the last added integer can cover it; if not, add two integers from the end of the current interval.
4Step 4: Continue until all intervals are covered.

solution.py15 lines

1# Full working Python code
2
3def minSetSize(intervals):
4    intervals.sort(key=lambda x: x[1])
5    nums = []
6    last_added = -1
7    for start, end in intervals:
8        if last_added < start:
9            nums.append(end - 1)
10            nums.append(end)
11            last_added = end
12        elif last_added == start:
13            nums.append(end)
14            last_added = end
15    return len(nums)

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) because we store the integers in a list.

1Greedy algorithms often provide efficient solutions for covering problems.
2Sorting the intervals helps in systematically choosing the best integers to cover multiple intervals.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.