LeetCode 757: Set Intersection Size At Least Two Solution
Master LeetCode problem 757 (Set Intersection Size At Least Two), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
Problem Explanation
Explanation:
To find the minimum possible size of a containing set, we can approach this problem greedily. We sort the intervals based on their end points in ascending order. Then, we iterate through the intervals and maintain two variables prev and curr to represent the last two numbers that are added to the containing set.
For each interval, we check if the current interval overlaps with the last two numbers in the set. If it does, we don't need to add any new numbers to the set. If it doesn't overlap, we add the current interval's end point and the end point - 1 to the set.
Time Complexity:
The time complexity of this algorithm is O(nlog(n)) where n is the number of intervals. The sorting step takes O(nlog(n)) time, and the iteration through the intervals takes O(n) time.
Space Complexity:
The space complexity of this algorithm is O(1) as we only use a constant amount of extra space.
Solution Code
class Solution {
public int intersectionSizeTwo(int[][] intervals) {
Arrays.sort(intervals, (a, b) -> a[1] - b[1]);
int prev = -1;
int curr = -1;
int count = 0;
for (int[] interval : intervals) {
if (interval[0] > curr) {
prev = interval[1] - 1;
curr = interval[1];
count += 2;
} else if (interval[0] > prev) {
prev = curr;
curr = interval[1];
count++;
}
}
return count;
}
}Try It Yourself
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