What is the brute force approach and complexity for Remove Covered Intervals?

For each interval, look for any other interval that contains it with time complexity O(N²) and space complexity O(1).

What is the optimal approach and complexity for Remove Covered Intervals?

Sort so earlier intervals start first; an interval survives only if it ends past maxEnd with optimal time complexity O(N log N) and space complexity O(1).

How do you identify the pattern for Remove Covered Intervals?

Look for key signals in the prompt: "list of intervals" indicates Sort first — by start is the default for this sub-pattern.; "covered: c ≤ a and b ≤ d" indicates A containment test, not an overlap test. Sorting by start handles c ≤ a; a running max handles b ≤ d.; "how many remain" indicates We only count; no need to build the merged list..

Remove Covered Intervals

Medium·LeetCode #1288 Pro OnlySlight Variation

LeetCode

Problem Statement

You get a list of intervals [l, r]. No interval appears twice.

Interval [a, b] is covered by interval [c, d] when it sits completely inside it: c ≤ a and b ≤ d.

Remove every interval that is covered by some other interval. Return how many intervals are left.

Input

A list intervals, where intervals[i] = [l, r].

Output

One integer: the number of intervals that are not covered by any other interval.

Example 1

Input: intervals = [[1,4],[3,6],[2,8]]

Output: 2

[3,6] is inside [2,8], so it is removed. [1,4] is not inside [2,8] because it starts earlier. 2 intervals remain.

Example 2

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

Output: 1

[2,3] is inside [1,4].

Constraints

• 1 ≤ intervals.length ≤ 1000
• intervals[i].length == 2
• 0 ≤ l < r ≤ 10^5
• All intervals are different.

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“list of intervals”

→ Sort first — by start is the default for this sub-pattern.

“covered: c ≤ a and b ≤ d”

→ A containment test, not an overlap test. Sorting by start handles c ≤ a; a running max handles b ≤ d.

“how many remain”

→ We only count; no need to build the merged list.

Approach Complexity Ladder

Brute Force — check every pairO(N²)O(1)

For each interval, look for any other interval that contains it

Optimal — sort by start (ties: end descending), track maxEndO(N log N)O(1)

Sort so earlier intervals start first; an interval survives only if it ends past maxEnd

Brute Force Intuition

For each interval i, loop over every other interval j. If j starts at or before i and ends at or after i, then i is covered. Count the intervals for which no such j exists.

Recognition rule for this sub-pattern:

Sort by start. If next.start ≤ current.end, they overlap — extend current.end. Otherwise start a new interval.

Remember this · Intervals

Trigger

Input is [start, end] pairs; overlap, merge, insert, meeting rooms, free time, minimum removals.

Think

After sorting, do I only ever need to compare against the most recent interval?

Key question

Sort by start or by end — which one makes the greedy choice obvious?

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