Divide Intervals Into Minimum Number of Groups
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Topics:
ArraysGreedy Algorithms
You are given a 2D integer array intervals where intervals[i] = [lefti, righti] represents the inclusive interval [lefti, righti].
You have to divide the intervals into one or more groups such that each interval is in exactly one group, and no two intervals that are in the same group intersect each other.
Return the minimum number of groups you need to make.
Two intervals intersect if there is at least one common number between them. For example, the intervals [1, 5] and [5, 8] intersect.
Example 1:
Input: intervals = [[5,10],[6,8],[1,5],[2,3],[1,10]] Output: 3 Explanation: We can divide the intervals into the following groups: - Group 1: [1, 5], [6, 8]. - Group 2: [2, 3], [5, 10]. - Group 3: [1, 10]. It can be proven that it is not possible to divide the intervals into fewer than 3 groups.
Example 2:
Input: intervals = [[1,3],[5,6],[8,10],[11,13]] Output: 1 Explanation: None of the intervals overlap, so we can put all of them in one group.
Constraints:
1 <= intervals.length <= 105intervals[i].length == 21 <= lefti <= righti <= 106
Solution
Clarifying Questions
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
- What are the possible ranges for the start and end values of the intervals? Are they integers?
- Can intervals overlap, be adjacent, or be identical (same start and end)?
- If the input list of intervals is empty or null, what should the function return?
- Is the input list of intervals sorted in any way, and if not, is there a preferred way to order them before processing?
- Do the intervals need to be grouped based on the smallest possible number of groups, or is an approximation sufficient?
Brute Force Solution
Approach
The brute force method for grouping intervals involves exploring every single possible combination of groups. We'll try every way to put each interval into a group, checking if that grouping is valid. We'll keep track of the best grouping we find that minimizes the number of groups needed.
Here's how the algorithm would work step-by-step:
- Start by considering the first interval. Try putting it into its own group.
- Next, consider the second interval. Try putting it into the same group as the first interval, and then try putting it into a new, separate group.
- For each subsequent interval, try putting it into every existing group and also into a new group. Check if the interval overlaps with any other intervals in the group it's being added to. If it does overlap, the grouping is invalid, and you must move on.
- Keep track of the number of groups required for each valid grouping of intervals.
- Once you have explored all possible groupings of all intervals, select the grouping that used the fewest number of groups. This is your answer.
Code Implementation
def divide_intervals_brute_force(intervals):
min_groups_needed = float('inf')
def overlap(interval1, interval2):
return not (interval1[1] < interval2[0] or interval2[1] < interval1[0])
def find_min_groups(index, current_groups):
nonlocal min_groups_needed
if index == len(intervals):
min_groups_needed = min(min_groups_needed, len(current_groups))
return
# Try placing the current interval in each existing group
for group_index in range(len(current_groups)):
valid_placement = True
for existing_interval in current_groups[group_index]:
if overlap(intervals[index], existing_interval):
valid_placement = False
break
if valid_placement:
current_groups[group_index].append(intervals[index])
find_min_groups(index + 1, current_groups)
current_groups[group_index].pop()
# If no existing group works, create a new group.
# Explore adding the interval to a NEW group.
find_min_groups(index + 1, current_groups + [[intervals[index]]])
# Start the recursion with an empty list of groups.
# Empty list simulates starting with no groups assigned.
find_min_groups(0, [])
return min_groups_needed
Big(O) Analysis
Time Complexity
O(k^n) – The brute force method explores every possible grouping of intervals. For each of the 'n' intervals, we consider placing it into each of the existing groups, or a new group. In the worst-case scenario, where no intervals overlap, each interval could potentially be placed into its own group. In this situation, the number of groups we can have can be as large as n groups. As a result, the number of possible groupings grows exponentially with the number of intervals. If we represent the maximum number of possible groups as k, for each of the n intervals, we may be choosing to add it to one of k possible groups, which results in approximately O(k^n) time complexity, where k can be <= n.
Space Complexity
O(N) – The brute force algorithm described explores all possible groupings of intervals, effectively building a decision tree where each level represents an interval and each branch represents assigning that interval to an existing or new group. The maximum depth of this tree is N, where N is the number of intervals. In the worst-case scenario, we need to keep track of the current grouping which consists of at most N groups, each potentially containing multiple intervals. Therefore, the space complexity is driven by storing this grouping information, leading to O(N) auxiliary space to track the current groups of intervals.
Optimal Solution
Approach
The core idea is to process the intervals in a specific order to minimize the number of groups needed. We sort the intervals and then efficiently assign them to groups, reusing groups whenever possible.
Here's how the algorithm would work step-by-step:
- First, organize the intervals by their starting points, putting the intervals that start earliest at the beginning.
- Next, create a structure to represent groups. Think of it like a list of meeting rooms, where each meeting room contains meetings that don't overlap.
- Now, go through each interval one by one, in the sorted order.
- For each interval, check if it can fit into any of the existing groups without overlapping. If it can, place it in the earliest available group.
- If the interval cannot fit into any existing group (because it overlaps with all of them), create a new group and place the interval in that new group.
- By assigning intervals in this way, you reuse existing groups as much as possible before creating new ones, which ensures the minimum number of groups is used.
Code Implementation
def divide_intervals_into_minimum_number_of_groups(intervals):
intervals.sort(key=lambda interval: interval[0])
groups = []
for interval in intervals:
# Iterate through each interval to assign to a group
interval_start = interval[0]
interval_end = interval[1]
assigned_to_group = False
for i in range(len(groups)):
last_interval_end = groups[i][-1][1]
# Check for overlaps before assigning to existing group
if interval_start > last_interval_end:
groups[i].append(interval)
assigned_to_group = True
break
if not assigned_to_group:
# No suitable group found, so creating a new group
groups.append([interval])
return len(groups)Big(O) Analysis
Time Complexity
O(n log n) – The dominant operations are sorting the intervals and finding a suitable group for each interval. Sorting n intervals takes O(n log n) time. For each interval, we iterate through the existing groups to find a non-overlapping group. In the worst case, we might need to check all existing groups. Since the number of groups is at most n, this check could take O(n) time per interval. However, if we use a priority queue to keep track of the end times of the intervals in each group (or even just sort the end times of the intervals in each group), we can find the first available group in O(log n) time. Thus, the overall time complexity is dominated by the sorting and becomes O(n log n) + n * O(log n), which simplifies to O(n log n).
Space Complexity
O(N) – The algorithm creates a data structure to represent the groups, which in the worst case, where all intervals are overlapping, could contain N groups. Each group will store one interval. Therefore, in the worst-case scenario where no intervals can be placed into existing groups, we might need N such groups, leading to O(N) auxiliary space. Input size, N represents the total number of intervals.
Edge Cases
| Case | How to Handle |
|---|---|
| Null or empty input intervals array | Return 0 as no groups are needed when there are no intervals. |
| Single interval in the input array | Return 1 since a single interval always requires one group. |
| Intervals with identical start and end times (e.g., [1,1]) | The algorithm should correctly process these as valid intervals needing a group. |
| Overlapping intervals that span a very large range of numbers (potential integer overflow if calculating sizes) | Use non-size-based overlapping checks or long integers if size is directly calculated and could cause overflow. |
| Very large number of intervals leading to memory constraints | Consider streaming or out-of-core algorithms if the input data is too large to fit into memory. |
| Intervals are already sorted or nearly sorted | The solution's time complexity should not degrade significantly if intervals are pre-sorted. |
| Intervals are reverse sorted | The solution must correctly handle reverse sorted intervals, possibly requiring sorting as part of the algorithm. |
| All intervals are non-overlapping | The algorithm should correctly identify that each interval needs its own group, resulting in a number of groups equal to the number of intervals. |
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