Non Overlapping Intervals Minimum Removal in DSA Python - Build from Scratch

Non Overlapping Intervals Minimum Removal

📖 Scenario: You are organizing a conference with multiple talks scheduled as time intervals. Some talks overlap, and you want to remove the minimum number of talks so that no two talks overlap.

🎯 Goal: Build a program that finds the minimum number of intervals to remove to make the rest non-overlapping.

📋 What You'll Learn

Create a list of intervals with exact values

Add a variable to count removals

Implement logic to find minimum removals for non-overlapping intervals

Print the minimum number of intervals to remove

💡 Why This Matters

🌍 Real World

Scheduling talks, meetings, or tasks without conflicts is common in event planning and project management.

💼 Career

Understanding interval scheduling and greedy algorithms is useful for software engineers working on calendar apps, resource allocation, and optimization problems.

1

Create the list of intervals

Create a list called intervals with these exact intervals: [1, 3], [2, 4], [3, 5], [6, 7], [8, 10], [9, 11]

DSA Python

# Your code here

Hint

Use a list of lists to represent intervals.

2

Add a variable to count removals

Add a variable called removals and set it to 0 to count how many intervals we remove

DSA Python

intervals = [[1, 3], [2, 4], [3, 5], [6, 7], [8, 10], [9, 11]]
# Your code here

Hint

Initialize removals to zero before processing intervals.

3

Implement logic to find minimum removals

Sort intervals by their end time using intervals.sort(key=lambda x: x[1]). Then create a variable prev_end set to the end of the first interval. Use a for loop with variable i from 1 to len(intervals) to check if the start of the current interval is less than prev_end. If yes, increment removals. Otherwise, update prev_end to the current interval's end.

DSA Python

intervals = [[1, 3], [2, 4], [3, 5], [6, 7], [8, 10], [9, 11]]
removals = 0
# Your code here

Hint

Sort intervals by end time to greedily keep intervals that finish earliest.

4

Print the minimum number of intervals to remove

Print the value of removals using print(removals)

DSA Python

intervals = [[1, 3], [2, 4], [3, 5], [6, 7], [8, 10], [9, 11]]
removals = 0
intervals.sort(key=lambda x: x[1])
prev_end = intervals[0][1]
for i in range(1, len(intervals)):
    if intervals[i][0] < prev_end:
        removals += 1
    else:
        prev_end = intervals[i][1]
# Your code here

Hint

The output is the minimum number of intervals to remove to avoid overlaps.

Practice

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

easy

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

B. To find the maximum number of overlapping intervals

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 A

4. Quick Check:

   Minimum removals = non-overlapping intervals [OK]

Hint: Focus on removing intervals to avoid overlaps, not merging [OK]

Common Mistakes:

• Confusing with merging intervals problem
• Trying to maximize overlaps instead of minimizing removals
• Sorting intervals incorrectly

2. Which of the following Python code snippets correctly sorts a list of intervals by their end time?

easy

A. intervals.sort(key=lambda x: x[0])

B. intervals.sort(reverse=True)

C. intervals.sort(key=lambda x: x[1])

D. intervals.sort(key=lambda x: x[1], reverse=True)

Solution

1. Step 1: Identify sorting by end time

   Sorting by end time means using the second element of each interval, which is index 1.

2. Step 2: Check the correct syntax

   Using intervals.sort(key=lambda x: x[1]) sorts intervals ascending by end time.

3. Final Answer:

   intervals.sort(key=lambda x: x[1]) -> Option C

4. Quick Check:

   Sort by end time = key=lambda x: x[1] [OK]

Hint: Sort intervals by end time using key=lambda x: x[1] [OK]

Common Mistakes:

• Sorting by start time instead of end time
• Using reverse=True when not needed
• Incorrect lambda syntax

3. What is the output of the following code?

intervals = [[1,3],[2,4],[3,5]]
intervals.sort(key=lambda x: x[1])
count = 0
end = float('-inf')
for interval in intervals:
    if interval[0] >= end:
        end = interval[1]
    else:
        count += 1
print(count)

medium

A. 2

B. 1

C. 0

D. 3

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals: [[1,3],[2,4],[3,5]] (already sorted by end time)

2. Step 2: Iterate and count overlaps

   Start with end = -inf
   Interval [1,3]: 1 >= -inf true, update end=3
   Interval [2,4]: 2 >= 3 false, count=1
   Interval [3,5]: 3 >= 3 true, update end=5

3. Final Answer:

   1 -> Option B

4. Quick Check:

   One interval removed to avoid overlap [OK]

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

Common Mistakes:

• Not updating end correctly
• Counting non-overlapping intervals instead of removals
• Sorting by start time instead of end time

4. Find the error in this code snippet that tries to solve the minimum removal of overlapping intervals:

def eraseOverlapIntervals(intervals):
    intervals.sort(key=lambda x: x[0])
    count = 0
    end = float('-inf')
    for interval in intervals:
        if interval[0] >= end:
            end = interval[1]
        else:
            count += 1
    return count

medium

A. The function should return intervals, not count

B. Using float('-inf') for end causes errors

C. The count variable should be initialized to 1

D. Sorting by start time instead of end time causes incorrect results

Solution

1. Step 1: Check sorting key

   The code sorts intervals by start time, but to minimize removals, sorting by end time is needed.

2. Step 2: Understand impact of sorting

   Sorting by start time can cause wrong overlap checks and incorrect count of removals.

3. Final Answer:

   Sorting by start time instead of end time causes incorrect results -> Option D

4. Quick Check:

   Sort by end time to minimize removals [OK]

Hint: Always sort intervals by end time for this problem [OK]

Common Mistakes:

• Sorting by start time
• Incorrect initialization of end
• Returning wrong value

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

   Start end = -inf
   [1,2]: 1 >= -inf true, end=2
   [2,3]: 2 >= 2 true, end=3
   [1,3]: 1 >= 3 false, count=1 (remove this)
   [3,4]: 3 >= 3 true, end=4

3. Final Answer:

   1 -> Option A

4. Quick Check:

   One interval removed to avoid overlap [OK]

Hint: Sort by end time, greedily pick intervals, count removals [OK]

Common Mistakes:

• Not sorting by end time
• Counting non-overlapping intervals instead of removals
• Removing wrong intervals