Why Non Overlapping Intervals Minimum Removal in DSA C++?

The Big Idea

What if you could fix a messy schedule by removing the fewest meetings possible, without checking every clash manually?

The Scenario

Imagine you have a list of meetings scheduled in a single room, but some meetings overlap in time. You want to remove the fewest meetings possible so that no two meetings clash.

The Problem

Trying to check every meeting against all others manually is slow and confusing. It's easy to miss overlaps or remove too many meetings, causing wasted time and frustration.

The Solution

This problem can be solved by sorting the meetings by their end times and then greedily choosing meetings that don't overlap. This way, you remove the minimum number of meetings to avoid clashes efficiently.

Before vs After

✗ Before

for (int i = 0; i < n; ++i) {
  for (int j = i + 1; j < n; ++j) {
    if (intervals[i] overlaps intervals[j]) {
      remove one meeting
    }
  }
}

✓ After

sort(intervals.begin(), intervals.end(), [](const Interval &a, const Interval &b) { return a.end < b.end; });
int count = 0;
int end = intervals[0].end;
for (int i = 1; i < n; ++i) {
  if (intervals[i].start < end) count++;
  else end = intervals[i].end;
}

What It Enables

This approach lets you quickly find the smallest number of meetings to cancel so that all remaining meetings fit without overlapping.

Real Life Example

Scheduling conference rooms in an office where meetings must not overlap, and you want to cancel as few meetings as possible to make the schedule work.

Key Takeaways

Manual checking of overlaps is slow and error-prone.

Sorting by end time and greedy selection solves the problem efficiently.

Enables minimal removal of intervals to avoid conflicts.

Practice

1. What is the main goal of the Non Overlapping Intervals Minimum Removal problem?

easy

A. To merge all intervals into one

B. To find the maximum number of overlapping intervals

C. To remove the minimum number of intervals so that the remaining intervals do not overlap

D. To sort intervals by their start time

Solution

1. Step 1: Understand the problem goal

   The problem asks to remove as few intervals as possible so that no two intervals overlap.

2. Step 2: Clarify what non-overlapping means

   Non-overlapping means for any two intervals, one ends before the other starts.

3. Final Answer:

   To remove the minimum number of intervals so that the remaining intervals do not overlap -> Option C

4. Quick Check:

   Minimum removal = non-overlapping intervals [OK]

Hint: Focus on removing intervals, not merging or counting overlaps [OK]

Common Mistakes:

• Confusing removal with merging intervals
• Thinking the goal is to count overlaps
• Sorting intervals without removing

2. Which of the following C++ code snippets correctly sorts intervals by their end time for the minimum removal problem?

easy

A. std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.end < b.end; });

B. std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.start < b.start; });

C. std::sort(intervals.begin(), intervals.end());

D. std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.end > b.end; });

Solution

1. Step 1: Identify sorting criteria for minimum removal

   We sort intervals by their end time ascending to greedily pick intervals that finish earliest.

2. Step 2: Check code correctness

   std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.end < b.end; }); uses a lambda comparing a.end < b.end, which is correct for sorting by end time ascending.

3. Final Answer:

   std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.end < b.end; }); -> Option A

4. Quick Check:

   Sort by end time ascending = std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.end < b.end; }); [OK]

Hint: Sort intervals by end time ascending for greedy approach [OK]

Common Mistakes:

• Sorting by start time instead of end time
• Sorting in descending order of end time
• Using default sort without comparator

3. What is the output of the following C++ code snippet?

struct Interval { int start, end; };
std::vector<Interval> intervals = {{1,3}, {2,4}, {3,5}};
std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.end < b.end; });
int count = 0, end_time = INT_MIN;
for (auto &interval : intervals) {
  if (interval.start >= end_time) {
    end_time = interval.end;
  } else {
    count++;
  }
}
std::cout << count << std::endl;

medium

A. 2

B. 1

C. 0

D. 3

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals by end: (1,3), (2,4), (3,5)

2. Step 2: Iterate and count removals

   Start with end_time = -∞ - Interval (1,3): start 1 >= -∞, keep, end_time=3 - Interval (2,4): start 2 < 3, overlap, count=1 - Interval (3,5): start 3 >= 3, keep, end_time=5

3. Final Answer:

   1 -> Option B

4. Quick Check:

   One overlap removed = 1 [OK]

Hint: Count overlaps by comparing start with last end time [OK]

Common Mistakes:

• Not updating end_time correctly
• Counting non-overlapping intervals as removals
• Sorting incorrectly before iteration

4. Identify the bug in the following C++ code for minimum interval removal:

std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.start < b.start; });
int count = 0, end_time = INT_MIN;
for (auto &interval : intervals) {
  if (interval.start >= end_time) {
    end_time = interval.end;
  } else {
    count++;
  }
}
return count;

medium

A. Not initializing count to 1 causes off-by-one error

B. Using INT_MIN for end_time causes overflow

C. Missing check for empty intervals vector

D. Sorting by start time instead of end time causes incorrect removal count

Solution

1. Step 1: Check sorting criteria

   The code sorts intervals by start time, but greedy approach requires sorting by end time to minimize removals.

2. Step 2: Understand impact on counting

   Sorting by start time can cause counting more removals than needed because intervals ending earlier are not prioritized.

3. Final Answer:

   Sorting by start time instead of end time causes incorrect removal count -> Option D

4. Quick Check:

   Sort by end time, not start time [OK]

Hint: Always sort by end time for minimum removal [OK]

Common Mistakes:

• Sorting by start time instead of end time
• Assuming INT_MIN causes overflow here
• Ignoring empty input checks

5. Given intervals [[1,2],[2,3],[3,4],[1,3]], what is the minimum number of intervals to remove to make the rest non-overlapping?

hard

A. 1

B. 2

C. 3

D. 0

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals: [1,2], [2,3], [1,3], [3,4]

2. Step 2: Select intervals greedily and count removals

   Keep [1,2] (end=2) - [2,3]: start 2 >= 2, keep - [1,3]: start 1 < 2 (overlaps with [1,2]), remove (count=1) - [3,4]: start 3 >= 3, keep Total removals so far: 1 But [1,3] overlaps with [1,2], removing it once is enough. Check if any other removals needed: none Total removals = 1

3. Step 3: Re-examine intervals order

   Actually, sorting by end time ascending is: [1,2], [2,3], [1,3], [3,4] But [1,3] overlaps with [1,2], so removing it once suffices. If we keep [1,2], [2,3], [3,4], we remove [1,3] only once. So minimum removals = 1

4. Final Answer:

   1 -> Option A

5. Quick Check:

   Remove overlapping [1,3] only once [OK]

Hint: Sort by end time, greedily pick intervals, count removals [OK]

Common Mistakes:

• Removing too many intervals unnecessarily
• Not sorting by end time
• Confusing overlapping intervals