Divide Intervals Into Minimum Number of Groups

class Solution {
public:
    int minGroups(vector<vector<int>>& intervals) {
        // Determine smallest and largest interval bounds
        int minIntervalStart = INT_MAX;
        int maxIntervalEnd = INT_MIN;
        for (vector<int> interval : intervals) {
            minIntervalStart = min(minIntervalStart, interval[0]);
            maxIntervalEnd = max(maxIntervalEnd, interval[1]);
        }

        // Build a difference array for interval counts
        vector<int> countDifferences(maxIntervalEnd + 2, 0);
        for (vector<int> interval : intervals) {
            countDifferences[interval[0]]++;
            countDifferences[interval[1] + 1]--;
        }

        int activeIntervals = 0;
        int maximumOverlap = 0;
        for (int i = minIntervalStart; i <= maxIntervalEnd; i++) {
            // Compute current active intervals
            activeIntervals += countDifferences[i];
            maximumOverlap = max(maximumOverlap, activeIntervals);
        }

        return maximumOverlap;
    }
};

This solution implements a method in C++ to determine the minimum number of groups needed to arrange a set of given intervals, such that no two intervals in the same group overlap. Here's a breakdown of how the solution works:

• First, determine the smallest and largest interval bounds across all intervals. This helps in setting the limits for building the difference array later.

• Create a difference array where each index corresponds to a point in time between the smallest start and the largest end of the intervals. An increment at a start point and a decrement after an end point are made for each interval.

• Traverse through the difference array from the smallest start to largest end, maintaining a count of 'active intervals' at each point by accumulating the differences. Track the maximum number of active intervals seen at any point, which essentially represents the maximum overlap.

• This maximum overlap hints at the minimum number of groups required, as each group can only contain non-overlapping intervals, and hence, during the time of maximum overlap, each interval would require a separate group.

The approach efficiently finds the answer by leveraging the difference array (also known as a sweep line technique), which allows for fast updates and queries about the number of overlapping intervals at each point in time.

class Solution {

    public int calculateMinGroups(int[][] ranges) {
        int minStart = Integer.MAX_VALUE;
        int maxEnd = Integer.MIN_VALUE;
        for (int[] range : ranges) {
            minStart = Math.min(minStart, range[0]);
            maxEnd = Math.max(maxEnd, range[1]);
        }

        int[] timeline = new int[maxEnd + 2];
        for (int[] range : ranges) {
            timeline[range[0]]++;
            timeline[range[1] + 1]--;
        }

        int currentActive = 0;
        int maxActive = 0;
        for (int i = minStart; i <= maxEnd; i++) {
            currentActive += timeline[i];
            maxActive = Math.max(maxActive, currentActive);
        }

        return maxActive;
    }
}

The provided Java code defines a method calculateMinGroups within the Solution class, aimed at solving the problem of dividing a set of intervals into the minimum number of groups so that no two intervals in the same group overlap. Here’s how the solution works:

• First, determine the minimum start and maximum end times of all intervals. This step helps in identifying the range over which to track interval overlaps.

• Create an array named timeline to keep count of active intervals at each point in time. This array is indexed from the smallest starting point to the greatest ending point (plus one for boundary handling).

• Iterate over the intervals. For each interval, increment the start time in timeline to denote the beginning of an interval and decrement the position right after the end time to denote the end of an interval.

  This annotating creates a map of how many intervals are active at each time point:

  • The value at timeline[start] is incremented, meaning a new interval starts; thus, increasing the count of active intervals.
  • The value at timeline[end + 1] is decremented, meaning an interval concludes; hence, decreasing the count of active intervals.

• Lastly, walk through the timeline array to calculate the total number of active intervals at any point in time. The peak of this count during the timeline traversal will be the answer, representing the minimum number of groups required.

This technique leverages a timeline sweeping strategy to efficiently count overlaps using linear time complexity relative to the number of time points considered.

This implementation effectively handles the described problem by using array manipulation to track active intervals, thereby calculating the required minimal grouping without examining each interval pair explicitly. The result is an optimal solution in terms of performance and clarity.

class Solution:
    def minOverlapIntervals(self, ranges: List[List[int]]) -> int:
        min_value = float("inf")
        max_value = float("-inf")
        for r in ranges:
            min_value = min(min_value, r[0])
            max_value = max(max_value, r[1])

        timeline = [0] * (max_value + 2)
        for r in ranges:
            timeline[r[0]] += 1
            timeline[r[1] + 1] -= 1

        current_overlap = 0
        max_overlap = 0
        for point in range(min_value, max_value + 1):
            current_overlap += timeline[point]
            max_overlap = max(max_overlap, current_overlap)

        return max_overlap

The Python function minOverlapIntervals is designed to find the minimum number of groups needed to divide given intervals so that no two intervals in the same group overlap. It uses a timeline sweeping algorithm for this purpose. Follow through to comprehend the efficiency and logic of the solution:

• Begin by determining the smallest and largest points across all the ranges. These represent the start and end of your timeline.

• A list named timeline is created to handle the increments and decrements at specific points in the timeline, which helps in understanding the number of intervals starting and ending at each point.

• For every interval in ranges, increase the value at the start index and decrease the value right after the end index in the timeline. This marks the span of the interval over the timeline by denoting where intervals begin and end.

• Traverse through the timeline from the minimum value to the maximum value to calculate overlapping intervals. Keep tracking the ongoing number of overlaps using current_overlap and update max_overlap whenever current_overlap exceeds the previous maximum.

• The result stored in max_overlap is the maximum number of overlapping intervals at any point, thus giving the minimum number of groups required to ensure no overlaps within each group.

This approach is efficient, leveraging the timeline strategy to handle overlapping intervals effectively with a straightforward implementation.