Non Overlapping Intervals Minimum Removal in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort intervals by their end time.

DSA C++

sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.end [1] b.end; });

Drag options to blanks, or click blank then click option'

A==

B>

C!=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' will sort in descending order, which is not helpful here.

Using '==' or '!=' will not sort properly.

Sorting intervals by their end time in ascending order helps to select intervals that finish earliest, minimizing removals.

2fill in blank

medium

Complete the code to check if the current interval overlaps with the last selected interval.

DSA C++

if (intervals[i].start [1] prev_end) { count++; } else { prev_end = intervals[i].end; }

Drag options to blanks, or click blank then click option'

A>=

B<=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>=' or '>' reverses the logic and causes wrong overlap detection.

Using '>' misses cases where intervals touch at endpoints.

If the current interval starts before the previous interval ends (start < prev_end), it overlaps. Using '<' checks for overlap.

3fill in blank

hard

Fix the error in the loop that iterates over intervals starting from the second element.

DSA C++

for (int i = [1]; i < intervals.size(); i++) { /* logic */ }

Drag options to blanks, or click blank then click option'

A1

Bintervals.size() - 1

C0

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Starting at 0 causes double counting the first interval.

Starting at intervals.size() - 1 or 2 skips necessary intervals.

We start from the second interval (index 1) because the first interval is always selected initially.

4fill in blank

hard

Fill both blanks to update the count of removals and the end time of the last selected interval.

DSA C++

if (intervals[i].start [1] prev_end) { [2]++; } else { prev_end = intervals[i].end; }

Drag options to blanks, or click blank then click option'

A<

B>

Ccount

Dremovals

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' reverses overlap logic.

Incrementing wrong variable name causes errors.

If current interval starts before previous end, it overlaps, so increment count of removals.

5fill in blank

hard

Fill all three blanks to complete the function that returns the minimum number of intervals to remove.

DSA C++

int eraseOverlapIntervals(vector<Interval>& intervals) {
    if (intervals.empty()) return 0;
    sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.end [1] b.end; });
    int count = 0;
    int prev_end = intervals[0].end;
    for (int i = [2]; i < intervals.size(); i++) {
        if (intervals[i].start [3] prev_end) {
            count++;
        } else {
            prev_end = intervals[i].end;
        }
    }
    return count;
}

Drag options to blanks, or click blank then click option'

A<

B1

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' in sort or overlap check reverses logic.

Starting loop at 0 causes double counting.

Sort by end ascending ('<'), start loop at 1, and check if current start < prev_end to detect overlap.

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