Non Overlapping Intervals Minimum Removal in DSA Java - 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 ends, removing overlapping ones.

intermediate

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

Sorting by end time helps to keep intervals that finish earliest, maximizing the number of non-overlapping intervals and minimizing removals.

beginner

What is the time complexity of the typical solution for Non Overlapping Intervals Minimum Removal?

O(n log n) due to sorting the intervals, where n is the number of intervals.

beginner

In the context of this problem, what does it mean if two intervals overlap?

Two intervals overlap if the start time of one is less than the end time of the other and vice versa, meaning they share some common time.

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

A Remove intervals randomly

B Sort intervals by their end time

C Sort intervals by their start time

D Count all intervals

Sorting by end time helps to select intervals that finish earliest, which is key to minimizing removals.

If intervals are [[1,3], [2,4], [3,5]], how many intervals must be removed to avoid overlap?

A 0

B 3

C 2

D 1

Removing [2,4] leaves [1,3] and [3,5] which do not overlap.

What does the minimum removal count equal in terms of intervals?

A Total intervals minus maximum number of non-overlapping intervals

B Maximum number of overlapping intervals

C Total intervals

D Maximum number of intervals

Minimum removals = total intervals - maximum non-overlapping intervals.

Which data structure is most useful to store intervals for this problem?

A Array or list of pairs

B Stack

C Queue

D Hash map

Intervals are best stored as arrays or lists of start-end pairs.

What is the key condition to check if two intervals overlap?

A End times are equal

B Start of one > end of the other

C Start of one < end of the other and vice versa

D Start times are equal

Intervals overlap if the start of one is before the end of the other and vice versa.

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

Think about choosing intervals that finish earliest to keep more intervals.

Describe how to determine if two intervals overlap and why this matters for the problem.

Focus on the time ranges and their intersections.

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