What signals in a problem statement indicate the Merge overlaps sub-pattern?

"merge all overlapping intervals"; "insert a new interval and merge if needed"; "remove intervals covered by another one"; "count the days not covered by any meeting"

Merge overlaps (Intervals Sub-pattern)

Q: What makes Merge overlaps different from other sub-patterns?

**Merge Intervals** is the template. **Insert Interval** is already sorted, so skip the sort: copy what ends before, absorb what overlaps, copy the rest — O(N). **Remove Covered Intervals** sorts by start, and on equal starts by end **descending**, so the bigger interval comes first and covers the smaller. **Count Days Without Meetings** merges, then subtracts the covered days from the total (`end − start + 1` each, since days are inclusive).

Sort by start and build the answer left to right. Each interval either extends the last merged block or starts a new one.

Signals in the problem statement

›
"merge all overlapping intervals"
›
"insert a new interval and merge if needed"
›
"remove intervals covered by another one"
›
"count the days not covered by any meeting"

What makes it different

Merge Intervals is the template. Insert Interval is already sorted, so skip the sort: copy what ends before, absorb what overlaps, copy the rest — O(N). Remove Covered Intervals sorts by start, and on equal starts by end descending, so the bigger interval comes first and covers the smaller. Count Days Without Meetings merges, then subtracts the covered days from the total (end − start + 1 each, since days are inclusive).

Questions

ordered from “learn the idea” to “recognise it without help”

1
Merge Intervals
Basic Anchor·LeetCode #56
WalkthroughMedium
2
Insert Interval
Basic Reinforcement·LeetCode #57
WalkthroughMedium
3
Remove Covered Intervals
Slight Variation·LeetCode #1288
WalkthroughMedium
4
Count Days Without Meetings
Pattern Transfer·LeetCode #3169
WalkthroughMedium

Template

merge-template.cpp

// Merge Intervals: sort by start, then extend the last merged block or start a new one.
vector<vector<int>> merge(vector<vector<int>>& intervals) {
    sort(intervals.begin(), intervals.end());        // by start (then end)
    vector<vector<int>> res;
    for (auto& cur : intervals) {
        if (!res.empty() && cur[0] <= res.back()[1])      // overlaps the LAST merged block
            res.back()[1] = max(res.back()[1], cur[1]);   // max: cur may sit inside it
        else
            res.push_back(cur);                           // gap → new block
    }
    return res;
}

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?

Intervals overview Scheduling / meeting rooms