LeetCode 2406: Divide Intervals Into Minimum Number of Groups Solution

Master LeetCode problem 2406 (Divide Intervals Into Minimum Number of Groups), a medium 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.

2406. Divide Intervals Into Minimum Number of Groups

Medium

Array Two Pointers Greedy Sorting Heap (Priority Queue)Prefix Sum

Problem 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.

Solution Code

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;
    }
}

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