Non Overlapping Intervals Minimum Removal in DSA Python

Mental Model

We want to remove the fewest intervals so that no intervals overlap. We do this by always keeping the interval that ends earliest to leave room for others.

Analogy: Imagine scheduling meetings in one room. To fit the most meetings without clashes, always pick the meeting that finishes earliest, then schedule the next meeting that starts after it ends.

Intervals: [1,3] -> [2,4] -> [3,5] -> null
Goal: Remove minimum intervals so none overlap.

Dry Run Walkthrough

Input: intervals: [[1,3], [2,4], [3,5]]

Goal: Remove the fewest intervals so that remaining intervals do not overlap

Step 1: Sort intervals by end time

[1,3] -> [2,4] -> [3,5]

Why: Sorting by end time helps pick intervals that finish earliest first

Step 2: Keep first interval [1,3], set end = 3

[1,3] ↑ -> [2,4] -> [3,5]

Why: Start with earliest finishing interval to maximize room for others

Step 3: Check next interval [2,4], it starts before end=3, so it overlaps; remove it

[1,3] ↑ -> [2,4](removed) -> [3,5]

Why: Overlap means we must remove this interval to avoid clash

Step 4: Check next interval [3,5], it starts at end=3, no overlap; keep it and update end=5

[1,3] -> [3,5] ↑

Why: No overlap, so we keep this interval and update end pointer

Result:

[1,3] -> [3,5] -> null
Minimum removals = 1

Annotated Code

DSA Python

from typing import List

class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        if not intervals:
            return 0
        # Sort intervals by their end time
        intervals.sort(key=lambda x: x[1])
        count = 0
        end = intervals[0][1]
        for i in range(1, len(intervals)):
            # If current interval starts before end, it overlaps
            if intervals[i][0] < end:
                count += 1  # Remove this interval
            else:
                end = intervals[i][1]  # Update end to current interval's end
        return count

# Driver code
intervals = [[1,3], [2,4], [3,5]]
sol = Solution()
print(sol.eraseOverlapIntervals(intervals))

intervals.sort(key=lambda x: x[1])

Sort intervals by their end time to pick earliest finishing intervals first

end = intervals[0][1]

Initialize end pointer to end of first interval

if intervals[i][0] < end:

Check if current interval overlaps with last kept interval

count += 1

Increment removal count for overlapping interval

else: end = intervals[i][1]

Update end pointer when interval does not overlap

Output

1

Complexity Analysis

Time: O(n log n) because we sort the intervals and then scan them once

Space: O(1) because sorting is in-place and only a few variables are used

vs Alternative: Naive approach checks all pairs for overlap leading to O(n^2) time, this greedy method is much faster

Edge Cases

empty intervals list

returns 0 because no intervals to remove

DSA Python

if not intervals:
    return 0

all intervals non-overlapping

returns 0 because no removals needed

DSA Python

if intervals[i][0] < end:

all intervals overlapping

removes all but one interval

DSA Python

count += 1

When to Use This Pattern

When asked to remove minimum intervals to avoid overlaps, use a greedy approach by sorting intervals by end time and removing overlapping ones.

Common Mistakes

Mistake: Sorting intervals by start time instead of end time
Fix: Sort intervals by their end time to ensure earliest finishing intervals are chosen first

Mistake: Not updating the end pointer after keeping an interval
Fix: Update end to current interval's end when interval does not overlap

Summary

Removes the fewest intervals so that no intervals overlap.

Use when you want to schedule intervals without clashes by removing minimum intervals.

Always pick intervals that finish earliest to leave room for others.