Leetcode #757: Set Intersection Size At Least Two
In this guide, we solve Leetcode #757 Set Intersection Size At Least Two in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 2D integer array intervals where intervals[i] = [starti, endi] represents all the integers from starti to endi inclusively. A containing set is an array nums where each interval from intervals has at least two integers in nums.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Greedy, Array, Sorting
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: intervals = [[1,3],[3,7],[8,9]]
Output: 5
Explanation: let nums = [2, 3, 4, 8, 9].
It can be shown that there cannot be any containing array of size 4.
Python Solution
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
Complexity
The time complexity is and the space complexity is , where is the number of intervals. The space complexity is , where is the number of intervals.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.