
Problem Statement
In the given problem, we are provided with an array of intervals, identified by intervals[i] = [starti, endi]
, where each interval denotes a start and an end point. The main objective here is to find out the minimum number of intervals that need to be removed such that the remaining intervals do not overlap each other. An important note is that two intervals that just touch at a boundary point, such as [1, 2]
and [2, 3]
, are considered non-overlapping. This problem is critical in scheduling, resource allocation, and various scenarios where conflict-free event planning is essential.
Examples
Example 1
Input:
intervals = [[1,2],[2,3],[3,4],[1,3]]
Output:
1
Explanation:
[1,3] can be removed and the rest of the intervals are non-overlapping.
Example 2
Input:
intervals = [[1,2],[1,2],[1,2]]
Output:
2
Explanation:
You need to remove two [1,2] to make the rest of the intervals non-overlapping.
Example 3
Input:
intervals = [[1,2],[2,3]]
Output:
0
Explanation:
You don't need to remove any of the intervals since they're already non-overlapping.
Constraints
1 <= intervals.length <= 105
intervals[i].length == 2
-5 * 104 <= starti < endi <= 5 * 104
Approach and Intuition
The problem can be approached by focusing on the strategy of minimizing the removal of intervals to achieve the highest number of non-overlapping intervals. Here is a breakdown of the intuitive steps:
First, sort the array of intervals based on their end times. Sorting by end time ensures that we always consider the earliest finishing time first, which maximizes the possibility of including subsequent intervals without overlap.
Initialize a variable, say
end
, to keep track of the end of the last added interval to the non-overlapping group. This variable helps in comparing with the start of the next interval in the sorted list.Iterate over the sorted intervals, and for each interval:
- Check if the start of the current interval is greater than or equal to the end of the last added interval.
- If true, push the current interval to the result as it does not overlap, and update the end variable to the end of this interval.
- If false, it means there is an overlap, and you skip adding this interval, effectively counting it as a removed one to minimize overlaps.
The count of skipped intervals gives the minimum number of removals needed.
These step-by-step logical applications facilitate the understanding of the approach used to minimize the number of interval overlaps by strategically removing the minimum number of intervals. This methodology not only ensures efficiency but also leverages sorting and greedy algorithms to simplify the problem-solving process.
Solutions
- C++
bool order_by_second(vector<int>& vec1, vector<int>& vec2) {
return vec1[1] < vec2[1];
}
class Solution {
public:
int eraseOverlapIntervals(vector<vector<int>>& intervals) {
sort(intervals.begin(), intervals.end(), order_by_second);
int count = 0;
int last_end = INT_MIN;
for (int i = 0; i < intervals.size(); i++) {
int start = intervals[i][0];
int end = intervals[i][1];
if (start >= last_end) {
last_end = end;
} else {
count++;
}
}
return count;
}
};
The provided C++ code offers a solution for finding the minimum number of intervals you need to remove from a set to ensure that no two intervals overlap. Here's a step-by-step explanation of what the code does:
The
order_by_second
function defines a custom comparator to sort intervals based on their endpoint values. It ensures that the vector of intervals will be sorted in ascending order by the second element of each sub-vector.Within the
Solution
class:The
eraseOverlapIntervals
function is defined to take a two-dimensional vector of integers, where each sub-vector represents an interval with a start and end point.The intervals are first sorted by their end times using the
sort()
function and the custom comparatororder_by_second
.Initialization is done for
count
which will keep a tally of the number of intervals that need to be removed, andlast_end
which tracks the end point of the last added interval, starting from the smallest possible integer.A loop iterates through each interval:
- The start and end of the current interval are determined.
- A check is performed to see if the current interval's start time is not less than the
last_end
. If true,last_end
is updated to the current interval's end time. Otherwise, it indicates an overlap, and the count is incremented.
After completing the loop through all intervals, the function returns the count of intervals that need to be removed.
This approach ensures an efficient determination of the minimum number of intervals to be removed for non-overlapping, employing sorting and a greedy algorithm strategy. The correctness is ensured as intervals are managed based on the least end time to maximize the chance of no overlap with forthcoming intervals.
- Java
class Solution {
public int removeOverlappingIntervals(int[][] data) {
Arrays.sort(data, Comparator.comparingInt(a -> a[1]));
int removed = 0;
int lastEnd = Integer.MIN_VALUE;
for (int i = 0; i < data.length; i++) {
int start = data[i][0];
int end = data[i][1];
if (start >= lastEnd) {
lastEnd = end;
} else {
removed++;
}
}
return removed;
}
}
The Java solution provided addresses the problem of finding the minimum number of intervals you need to remove to make the rest non-overlapping. It devices a strategic method by first sorting the intervals based on their end times, and then iterating through the sorted list to determine the necessity of removing an interval.
First, sort the array of intervals
data
based on the endpoint of each interval. This ensures that you always have the smallest possible 'end' to compare with, optimizing the chances of the subsequent interval not overlapping.Initialize an integer
removed
to count the number of intervals that need removal to prevent overlap.Initialize
lastEnd
to the smallest possible integer value. This variable keeps track of the endpoint of the last added interval that did not overlap.Iterate over each interval:
- Extract the start and end of the current interval.
- If the start of the current interval is greater than or equal to
lastEnd
, updatelastEnd
to this interval’s end since this interval does not overlap with the previously stored interval. - If the start is less than
lastEnd
, increment theremoved
counter as this interval overlaps with the previous one and needs to be removed.
Return the
removed
count which represents the minimum number of intervals to eliminate to ensure that none of the remaining intervals overlap.
By efficiently selecting only those intervals that finish the earliest and do not overlap, this solution minimizes the deletions necessary to achieve non-overlapping intervals.
- Python
class Solution:
def removeOverlapping(self, intervals: List[List[int]]) -> int:
intervals.sort(key=lambda interval: interval[1])
removed = 0
last_end = float('-inf')
for start, end in intervals:
if start >= last_end:
last_end = end
else:
removed += 1
return removed
In this Python solution, the goal is to determine the minimum number of intervals you need to remove to ensure that no intervals overlap.
Here's a step-by-step review of how the solution works:
Sort the
intervals
list based on the endpoint of each interval using a lambda function. This helps in selecting the intervals that finish the earliest, allowing more room for subsequent intervals to be included without overlapping.Initialize
removed
to 0, which will keep track of the count of intervals that need to be removed to avoid overlap.Initialize
last_end
to negative infinity. This variable tracks the end of the last interval that was added without overlapping.Iterate through each interval. For each
start
andend
:- If
start
is greater than or equal tolast_end
, updatelast_end
to the currentend
. This means the current interval does not overlap with the previously considered interval. - If
start
is less thanlast_end
, increment theremoved
counter by 1 because the current interval overlaps with the previous interval and must be removed to avoid overlap.
- If
Return the
removed
count, which is the number of intervals removed to achieve the necessary non-overlapping condition.
By adhering to this approach, the solution efficiently calculates the minimum number of deletions required to ensure that no two intervals overlap, using sorting and a simple iteration through the list.
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