Non Overlapping Intervals Minimum Removal in DSA Java - 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 of removals

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

Execution Sample

DSA Java

int eraseOverlapIntervals(int[][] intervals) {
  Arrays.sort(intervals, (a,b) -> a[1] - b[1]);
  int count = 0, end = Integer.MIN_VALUE;
  for (int[] interval : intervals) {
    if (interval[0] >= end) end = interval[1];
    else count++;
  }
  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	End Pointer	Count Removals	Visual State
1	Sort intervals	[1,2], [2,3], [3,4], [1,3]	N/A	0	[1,2] -> [1,3] -> [2,3] -> [3,4]
2	Initialize	N/A	-∞	0	Empty processed list
3	Check interval start >= end?	[1,2]	-∞	0	[1,2] selected
4	Update end	[1,2]	2	0	[1,2] selected
5	Check interval start >= end?	[1,3]	2	0	[1,2] selected
6	Remove interval	[1,3]	2	1	[1,2] selected; [1,3] removed
7	Check interval start >= end?	[2,3]	2	1	[1,2] selected; [1,3] removed
8	Update end	[2,3]	3	1	[1,2] -> [2,3] selected; [1,3] removed
9	Check interval start >= end?	[3,4]	3	1	[1,2] -> [2,3] selected; [1,3] removed
10	Update end	[3,4]	4	1	[1,2] -> [2,3] -> [3,4] selected; [1,3] removed
11	End loop	N/A	4	1	Final intervals: [1,2] -> [2,3] -> [3,4]
12	Return count	N/A	4	1	Minimum removals = 1

💡 All intervals processed; count of removals is 1 because [1,3] overlaps with [2,3]

Variable Tracker

Variable	Start	After Step 3	After Step 5	After Step 7	After Step 9	Final
end	-∞	2	2	3	4	4
count	0	0	1	1	1	1

Key Moments - 3 Insights

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

Sorting by end time ensures we always pick the interval that finishes earliest, leaving more room for following intervals. This is shown in execution_table step 1 where intervals are sorted by end time.

Why do we increase count when interval start is less than end?

If the current interval starts before the last selected interval ends, it overlaps and must be removed. This is shown in execution_table step 6 where count increments because [1,3] overlaps with the last end=2.

Why do we update 'end' only when interval start >= end?

We update 'end' to the current interval's end only if it doesn't overlap, meaning it can be part of the non-overlapping set. This is shown in steps 4, 8, and 10 where 'end' moves forward.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'count' after step 9?

A 0

B 2

C 1

D 3

💡 Hint

Check the 'Count Removals' column at step 9 in execution_table.

At which step does the algorithm decide to remove an interval?

A Step 10

B Step 6

C Step 8

D Step 12

💡 Hint

Look for 'Remove interval' operation in execution_table.

If the intervals were not sorted by end time, how would the 'end' pointer update change?

A It would update more frequently, possibly causing incorrect removals.

B It would never update.

C It would update only once at the start.

D It would update randomly without pattern.

💡 Hint

Refer to concept_flow and execution_table step 1 about sorting importance.

Concept Snapshot

Non Overlapping Intervals Minimum Removal:
- Sort intervals by end time.
- Initialize count=0, end=-∞.
- For each interval:
  - If start >= end, update end.
  - Else, increment count (remove interval).
- 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 earliest. We initialize a count of removals to zero and set an end pointer to negative infinity. Then, we iterate through each interval. If the current interval starts after or exactly when the last selected interval ends, we update the end pointer to this interval's end. Otherwise, the interval overlaps and must be removed, so we increment the count. After processing all intervals, the count represents the minimum removals needed. This approach ensures the maximum number of non-overlapping intervals are kept.

Practice

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

easy

A. To find the maximum number of overlapping intervals

B. To find the minimum number of intervals to remove so that the rest do not overlap

C. To sort intervals by their start time

D. To merge all overlapping intervals into one

Solution

1. Step 1: Understand the problem goal

   The problem asks to remove the fewest intervals so that no two intervals overlap.

2. Step 2: Differentiate from related problems

   Unlike merging intervals or counting overlaps, here we minimize removals to avoid overlaps.

3. Final Answer:

   To find the minimum number of intervals to remove so that the rest do not overlap -> Option B

4. Quick Check:

   Minimum removals = no overlaps [OK]

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

Common Mistakes:

• Confusing with merging intervals problem
• Thinking the goal is to count overlaps
• Assuming sorting alone solves the problem

2. Which of the following Java code snippets correctly sorts an array of intervals by their end time for solving the minimum removal problem?

easy

A. Arrays.sort(intervals, (a, b) -> a[0] - b[0]);

B. Arrays.sort(intervals, (a, b) -> b[1] - a[1]);

C. Arrays.sort(intervals, (a, b) -> a[1] - b[1]);

D. Arrays.sort(intervals);

Solution

1. Step 1: Identify sorting criteria for greedy approach

   We sort intervals by their end time ascending to pick earliest finishing intervals first.

2. Step 2: Match code to sorting by end time ascending

   Arrays.sort(intervals, (a, b) -> a[1] - b[1]); uses comparator (a, b) -> a[1] - b[1], which sorts by end time ascending.

3. Final Answer:

   Arrays.sort(intervals, (a, b) -> a[1] - b[1]); -> Option C

4. Quick Check:

   Sort by end time ascending = Arrays.sort(intervals, (a, b) -> a[1] - b[1]); [OK]

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

Common Mistakes:

• Sorting by start time instead of end time
• Sorting in descending order of end time
• Using default sort without comparator for 2D array

3. What is the output of the following Java code snippet that finds the minimum number of intervals to remove?

int[][] intervals = {{1,3},{2,4},{3,5}};
Arrays.sort(intervals, (a,b) -> a[1] - b[1]);
int count = 0, end = intervals[0][1];
for (int i = 1; i < intervals.length; i++) {
  if (intervals[i][0] < end) {
    count++;
  } else {
    end = intervals[i][1];
  }
}
System.out.println(count);

medium

A. 1

B. 0

C. 2

D. 3

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals: [1,3], [2,4], [3,5]

2. Step 2: Iterate and count overlaps

   Start with end=3. Check [2,4]: 2 < 3 overlap, count=1. Check [3,5]: 3 < 3 false, update end=5.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   One overlap removed = 1 [OK]

Hint: Count intervals starting before current end to find removals [OK]

Common Mistakes:

• Not updating end when no overlap
• Counting non-overlapping intervals as removals
• Incorrectly sorting intervals

4. Identify the bug in the following Java code that attempts to solve the minimum removal problem:

int[][] intervals = {{1,2},{2,3},{3,4},{1,3}};
Arrays.sort(intervals, (a,b) -> a[0] - b[0]);
int count = 0, end = intervals[0][1];
for (int i = 1; i < intervals.length; i++) {
  if (intervals[i][0] < end) {
    count++;
  } else {
    end = intervals[i][1];
  }
}
System.out.println(count);

medium

A. Missing check for empty intervals array

B. Not initializing count to 1 causes wrong output

C. Using < instead of <= in if condition causes errors

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.

2. Step 2: Understand impact on counting removals

   Sorting by start time can cause wrong overlap detection and incorrect removal count.

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 correct greedy removal count [OK]

Common Mistakes:

• Sorting by start time instead of end time
• Confusing overlap condition with equality
• Ignoring empty input edge case

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

hard

A. 2

B. 4

C. 3

D. 1

Solution

1. Step 1: Sort intervals by end time ascending

   Sorted: [1,2], [2,3], [1,3], [3,4], [2,5]

2. Step 2: Select intervals greedily and count removals

   Pick [1,2], end=2. [2,3] start=2 ≥ 2 no removal, end=3. [1,3] start=1 < 3 removal=1. [3,4] start=3 ≥ 3 no removal, end=4. [2,5] start=2 < 4 removal=2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Two removals needed = 2 [OK]

Hint: Sort by end time, pick earliest finishing intervals, count overlaps [OK]

Common Mistakes:

• Not sorting by end time before processing
• Counting intervals that do not overlap as removals
• Forgetting to update end after selecting interval