Minimize Connected Groups by Inserting Interval Solution in Python

C++
Java

Problem Description

Ah, the classic “Minimize Connected Groups by Inserting Interval” problem! It’s like trying to fit your entire wardrobe into a suitcase that’s just a tad too small. You know you have to make some sacrifices, but which clothes do you leave behind? In this problem, you’re given a bunch of intervals (think of them as your clothes) and a number k (the size of your suitcase). Your goal? Minimize the number of connected groups of intervals by inserting new intervals of size k.

Imagine you’re at a party, and you want to group your friends based on their interests. But wait! Some of them are too far apart in their interests, and you need to insert a few more friends (intervals) to make the groups work. It’s a social nightmare!

Code Solution


class Solution:
    def minConnectedGroups(self, intervals: list[list[int]], k: int) -> int:
        mergedIntervals = 0
        maxMergedIntervals = 0

        intervals = self._merge(intervals)

        i = 0
        for _, end in intervals:
            while i < len(intervals) and end + k >= intervals[i][0]:
                mergedIntervals += 1
                i += 1
            mergedIntervals -= 1  # Exclude intervals[i].
            maxMergedIntervals = max(maxMergedIntervals, mergedIntervals)

        return len(intervals) - maxMergedIntervals

    # Same as 56. Merge Intervals
    def _merge(self, intervals: list[list[int]]) -> list[list[int]]:
        res = []
        for interval in sorted(intervals):
            if not res or res[-1][1] < interval[0]:
                res.append(interval)
            else:
                res[-1][1] = max(res[-1][1], interval[1])
        return res

Approach

The approach taken in this solution is quite elegant. First, it merges overlapping intervals using a helper function _merge. Then, it iterates through the merged intervals to count how many can be connected by inserting new intervals of size k. The goal is to maximize the number of merged intervals, and the final answer is simply the total number of intervals minus the maximum number of merged intervals.

Time and Space Complexity

Complexity Type	Complexity
Time Complexity	O(n log n)
Space Complexity	O(n)

Real-World Example

Imagine you’re organizing a series of events (intervals) for a community festival. Each event has a start and end time, and you want to ensure that events that are close enough (within k hours) can be grouped together. However, some events are too far apart, and you need to insert new events to bridge the gaps. This problem helps you figure out the minimum number of groups you can create by strategically inserting new events.