Non Overlapping Intervals Minimum Removal in DSA C++ - Execution Trace

Concept Flow - Non Overlapping Intervals Minimum Removal

Sort intervals by end time

↓

Initialize count = 0, end = -∞

↓

For each interval

↓

Check if interval start >= end?

No→Remove interval, count++

Yes↓

Update end = interval end

↓

Repeat for all intervals

↓

Return count as minimum removals

Sort intervals by their end time, then iterate to count how many intervals must be removed to avoid overlaps.

Execution Sample

DSA C++

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

This code sorts intervals by end time and counts how many intervals overlap and must be removed.

Execution Table

Step	Operation	Current Interval	Condition (start < end?)	Action	Count	End	Intervals Removed
1	Sort intervals	[1,3], [2,4], [3,5]	N/A	Sorted by end time	0	-∞	[]
2	Check interval	[1,3]	1 < -∞? No	Update end = 3	0	3	[]
3	Check interval	[2,4]	2 < 3? Yes	Remove interval, count++	1	3	[[2,4]]
4	Check interval	[3,5]	3 < 3? No	Update end = 5	1	5	[[2,4]]
5	End	All intervals processed	N/A	Return count	1	5	[[2,4]]

💡 All intervals checked; minimum removals = 1

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	Final
count	0	0	1	1	1
end	-∞	3	3	5	5
intervals_removed	[]	[]	[[2,4]]	[[2,4]]	[[2,4]]

Key Moments - 3 Insights

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

Sorting by end time ensures we always keep the interval that finishes earliest, leaving more room for following intervals. See Step 1 in execution_table where sorting is done by end time.

Why do we compare interval start with 'end' and not interval end?

We compare the current interval's start with the last kept interval's end to detect overlap. If start < end, intervals overlap and removal is needed. See Steps 3 and 4 in execution_table.

Why do we update 'end' only when intervals do not overlap?

Updating 'end' to the current interval's end means we keep this interval. If intervals overlap, we do not update 'end' because we remove the current interval instead. See Steps 3 and 4.

Visual Quiz - 3 Questions

Look at the execution_table at Step 3, what is the value of 'count' after processing interval [2,4]?

A 1

B 2

C 0

D 3

💡 Hint

Check the 'Count' column at Step 3 in execution_table.

At which step does the condition 'start < end' become false for the last time?

A Step 3

B Step 4

C Step 2

D Step 5

💡 Hint

Look at the 'Condition (start < end?)' column in execution_table for each step.

If the interval [2,4] was changed to [4,6], how would the count change after Step 3?

A count would increase to 1

B count would increase to 2

C count would remain 0

D count would decrease to -1

💡 Hint

Compare the condition 'start < end' at Step 3 with the new interval start.

Concept Snapshot

Non Overlapping Intervals Minimum Removal:
- Sort intervals by end time.
- Initialize count=0, end=-∞.
- For each interval:
  - If start < end, increment count (remove interval).
  - Else update end = interval end.
- Return count as minimum removals.

Full Transcript

This concept solves the problem of removing the minimum number of intervals to avoid overlaps. First, intervals are sorted by their end times to prioritize intervals that finish earlier. Then, we iterate through each interval, checking if it overlaps with the last kept interval by comparing the current interval's start with the stored end. If it overlaps, we increase the removal count. Otherwise, we update the end to the current interval's end. The final count after processing all intervals is the minimum number of removals needed.

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