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, prev_end = -∞

↓

For each interval

↓

Check if current start >= prev_end?

No→Remove interval, count++

Yes↓

Update prev_end = current end

↓

Repeat for all intervals

↓

Return count

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

Execution Sample

DSA C

int cmp(const void *a, const void *b) {
  return ((int (*)[2])a)[0][1] - ((int (*)[2])b)[0][1];
}
int eraseOverlapIntervals(int intervals[][2], int n) {
  if (n <= 1) return 0;
  qsort(intervals, n, sizeof(intervals[0]), cmp);
  int count = 0, prev_end = intervals[0][1];
  for (int i = 1; i < n; i++) {
    if (intervals[i][0] < prev_end) count++;
    else prev_end = intervals[i][1];
  }
  return count;
}

This code sorts intervals by end time and counts minimum removals to avoid overlaps.

Execution Table

Step	Operation	Current Interval	Prev End	Overlap?	Count	Visual State
1	Sort intervals	[1,2], [2,3], [1,3], [3,4]	-	No	0	[1,2] -> [2,3] -> [1,3] -> [3,4]
2	Initialize	[1,2]	2	No	0	[1,2] -> [2,3] -> [1,3] -> [3,4]
3	Check interval	[2,3]	2	No	0	[1,2] -> [2,3] -> [1,3] -> [3,4]
4	Update prev_end	[2,3]	3	No	0	[1,2] -> [2,3] -> [1,3] -> [3,4]
5	Check interval	[1,3]	3	Yes	1	[1,2] -> [2,3] -> [1,3] (removed) -> [3,4]
6	Check interval	[3,4]	3	No	1	[1,2] -> [2,3] -> [1,3] (removed) -> [3,4]
7	Update prev_end	[3,4]	4	No	1	[1,2] -> [2,3] -> [3,4]
8	End	-	4	-	1	[1,2] -> [2,3] -> [3,4]

💡 All intervals checked; total removals = 1

Variable Tracker

Variable	Start	After Step 3	After Step 5	After Step 7	Final
count	0	0	1	1	1
prev_end	-	2	3	4	4

Key Moments - 3 Insights

Why do we update prev_end only when no overlap is found?

Because prev_end tracks the end of the last interval kept. If overlap occurs, we remove the current interval and do not update prev_end, as the previous interval remains.

Why do we sort intervals by their end time?

Sorting by end time ensures we always keep intervals that finish earliest, maximizing space for future intervals and minimizing removals, as shown in Step 1 of execution_table.

What happens if intervals have the same end time?

They are processed in order after sorting. Overlaps are detected by comparing start with prev_end, so intervals with same end time but overlapping starts will cause removals.

Visual Quiz - 3 Questions

Look at the execution_table at Step 7, what is the count value?

A 2

B 0

C 1

D 3

💡 Hint

Check the 'Count' column at Step 7 in execution_table.

At which step does prev_end update to 3?

A Step 4

B Step 6

C Step 5

D Step 3

💡 Hint

Look at the 'prev_end' column changes in execution_table rows.

If the first interval was [1,5] instead of [1,2], how would count change?

A It would stay the same

B It would increase

C It would decrease

D Cannot determine

💡 Hint

Consider how a longer first interval affects overlaps and removals in the execution_table.

Concept Snapshot

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

Full Transcript

To find the minimum number of intervals to remove so that the rest do not overlap, first sort intervals by their end times. Initialize a count of removals to zero and set prev_end to the end of the first interval. Iterate through each interval starting from the second. If the current interval's start is less than prev_end, it overlaps, so increment count and do not update prev_end. Otherwise, update prev_end to the current interval's end. After checking all intervals, the count represents the minimum removals needed.