Non Overlapping Intervals Minimum Removal in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

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

To find the minimum number of intervals to remove from a list so that the remaining intervals do not overlap.

intermediate

Which greedy strategy is commonly used to solve the Non Overlapping Intervals Minimum Removal problem?

Sort intervals by their end time and iteratively select intervals that start after the last selected interval's end to maximize non-overlapping intervals.

intermediate

Why do we sort intervals by their end time instead of start time in this problem?

Sorting by end time helps to always pick the interval that finishes earliest, leaving more room for the rest, thus maximizing the number of non-overlapping intervals.

beginner

What does the minimum removal count equal in terms of total intervals and maximum non-overlapping intervals?

Minimum removals = Total number of intervals - Maximum number of non-overlapping intervals.

beginner

In the C++ solution, what data structure is typically used to store intervals and why?

A vector of pairs or a vector of vectors is used to store intervals because it allows easy sorting and iteration.

What is the first step in solving the Non Overlapping Intervals Minimum Removal problem?

A Remove all intervals that overlap

B Sort intervals by their start time

C Sort intervals by their end time

D Count the total intervals

Sorting by end time helps to pick intervals that finish earliest, maximizing non-overlapping intervals.

If you have 5 intervals and the maximum number of non-overlapping intervals is 3, how many intervals must be removed?

A 2

B 3

C 5

D 0

Minimum removals = Total intervals (5) - Maximum non-overlapping (3) = 2.

Which of these intervals overlaps with [1, 3]?

A [2, 4]

B [3, 5]

C [4, 6]

D [0, 1]

Interval [2, 4] overlaps with [1, 3] because 2 < 3.

What is the time complexity of sorting intervals by their end time?

A O(log n)

B O(n log n)

C O(n)

D O(n^2)

Sorting n intervals takes O(n log n) time.

After sorting intervals by end time, how do you decide if an interval can be included in the non-overlapping set?

A If it has the longest duration

B If its end time is less than the start time of the last included interval

C If it overlaps with the first interval

D If its start time is greater or equal to the end time of the last included interval

An interval can be included if it starts after or exactly when the last included interval ends.

Explain the greedy approach to solve the Non Overlapping Intervals Minimum Removal problem step-by-step.

Think about how choosing intervals that finish earliest helps.

Describe how to determine if two intervals overlap and how this affects the solution.

Focus on interval start and end comparisons.

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