LeetCode 2406: Divide Intervals Into Minimum Number of Groups
LeetCode 2406 Solution Explanation
Explanation
To solve this problem, we can use the following algorithm:
- Sort the intervals based on their start points.
- Initialize an empty list to store the endpoints of the groups.
- Iterate through each interval in the sorted list.
- If the endpoint of the current interval is greater than all endpoints of the existing groups, add the endpoint to a new group.
- Otherwise, update the endpoint of the appropriate existing group to the endpoint of the current interval.
- The number of groups needed will be equal to the number of distinct endpoints.
Time complexity: O(n log n) where n is the number of intervals (due to sorting) Space complexity: O(n) for storing the sorted intervals and endpoints.
LeetCode 2406 Solutions in Java, C++, Python
import java.util.Arrays;
class Solution {
public int minGroups(int[][] intervals) {
Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
int groups = 0;
int[] endpoints = new int[intervals.length];
for (int[] interval : intervals) {
int i = 0;
while (i < groups && endpoints[i] < interval[0]) {
i++;
}
if (i == groups) {
endpoints[groups] = interval[1];
groups++;
} else {
endpoints[i] = Math.max(endpoints[i], interval[1]);
}
}
return groups;
}
}Interactive Code Editor for LeetCode 2406
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