Non Overlapping Intervals Minimum Removal in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

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

The goal is to find the minimum number of intervals to remove from a list so that the remaining intervals do not overlap.

intermediate

Why do we sort intervals by their end time in the Non Overlapping Intervals Minimum Removal problem?

Sorting by end time helps us greedily select intervals that finish earliest, leaving more room for others and minimizing removals.

beginner

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, meaning they share some common time.

intermediate

In the greedy approach, how do we decide which interval to keep when intervals overlap?

We keep the interval with the earliest end time and remove the others that overlap with it.

intermediate

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

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

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 overlapping intervals without sorting

Sorting intervals by their end time allows the greedy approach to select intervals that finish earliest, minimizing removals.

If intervals [1,3] and [2,4] overlap, which interval should be removed to minimize removals?

A Remove [1,3]

B Remove none

C Remove [2,4]

D Remove both

Keeping the interval with the earlier end time ([1,3]) allows more intervals to fit without overlap.

What does the minimum removal count represent in this problem?

A Number of overlapping pairs

B Number of intervals kept

C Total number of intervals

D Number of intervals removed to avoid overlap

The minimum removal count is how many intervals must be removed so that no intervals overlap.

Which algorithmic approach is commonly used to solve this problem efficiently?

A Greedy Algorithm

B Dynamic Programming

C Backtracking

D Divide and Conquer

A greedy algorithm that selects intervals by earliest end time is efficient and effective here.

What is the main reason for sorting intervals before processing?

A To process intervals in increasing order of end time

B To group overlapping intervals

C To find the longest interval

D To remove duplicates

Sorting by end time helps pick intervals that finish earliest, maximizing the number of non-overlapping intervals.

Explain the greedy strategy used to find the minimum number of intervals to remove to avoid overlaps.

Think about choosing intervals that finish earliest to leave room for others.

Describe how to detect if two intervals overlap and how that affects the removal decision.

Overlap means intervals share time; removing the one that ends later helps.

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 merge all intervals into one

D. To sort intervals by their start time

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: Clarify what non-overlapping means

   Non-overlapping intervals do not share any common points except possibly at the boundary.

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 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
• Assuming intervals can be rearranged

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

easy

A. qsort(intervals, n, sizeof(Interval), compareStart);

B. qsort(intervals, n, sizeof(Interval), compareEnd);

C. qsort(intervals, n, sizeof(Interval), compareLength);

D. qsort(intervals, n, sizeof(Interval), compareStartDesc);

Solution

1. Step 1: Identify sorting criteria for minimum removal

   Sorting intervals by their end time helps greedily select intervals that finish earliest.

2. Step 2: Match sorting function

   The function compareEnd sorts by end time, which is correct for this problem.

3. Final Answer:

   qsort(intervals, n, sizeof(Interval), compareEnd); -> Option B

4. Quick Check:

   Sort by end time = compareEnd [OK]

Hint: Sort intervals by end time to minimize removals [OK]

Common Mistakes:

• Sorting by start time instead of end time
• Sorting by interval length which is irrelevant
• Sorting in descending order of start time

3. What is the output of the following C code snippet that calculates minimum removals for intervals?

typedef struct { int start; int end; } Interval;

int minRemoval(Interval intervals[], int n) {
    qsort(intervals, n, sizeof(Interval), compareEnd);
    int count = 0, end = intervals[0].end;
    for (int i = 1; i < n; i++) {
        if (intervals[i].start < end) {
            count++;
        } else {
            end = intervals[i].end;
        }
    }
    return count;
}

// Intervals: {{1,3}, {2,4}, {3,5}}

int main() {
    Interval intervals[] = {{1,3}, {2,4}, {3,5}};
    int n = 3;
    printf("%d", minRemoval(intervals, n));
    return 0;
}

medium

A. 3

B. 2

C. 0

D. 1

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals by end: {1,3}, {2,4}, {3,5}

2. Step 2: Count overlaps and removals

   Start with end=3. Interval {2,4} starts at 2 < 3, so count=1. Interval {3,5} starts at 3 >= 3, update end=5.

3. Final Answer:

   1 -> Option D

4. Quick Check:

   One overlap removed = 1 [OK]

Hint: Count intervals starting before last end time [OK]

Common Mistakes:

• Not sorting intervals before counting
• Counting intervals that do not overlap
• Updating end incorrectly

4. Identify the bug in this C function for minimum interval removal:

int minRemoval(Interval intervals[], int n) {
    qsort(intervals, n, sizeof(Interval), compareStart);
    int count = 0, end = intervals[0].end;
    for (int i = 1; i < n; i++) {
        if (intervals[i].start < end) {
            count++;
        } else {
            end = intervals[i].end;
        }
    }
    return count;
}

medium

A. Using compareStart instead of compareEnd in qsort

B. Not initializing count to zero

C. Loop starts from 0 instead of 1

D. Updating end inside the if condition

Solution

1. Step 1: Check sorting criteria

   The function sorts intervals by start time, but it should sort by end time for correct greedy selection.

2. Step 2: Understand impact of wrong sorting

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

3. Final Answer:

   Using compareStart instead of compareEnd in qsort -> Option A

4. Quick Check:

   Sort by end time needed [OK]

Hint: Always sort by end time, not start time [OK]

Common Mistakes:

• Sorting by start time
• Incorrect loop index
• Wrong update of end variable

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. 3

B. 2

C. 1

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

   Select {1,2} first (end=2). {2,3} starts at 2 >= 2, select it (end=3). {1,3} starts at 1 < 3, remove it (count=1). {3,4} starts at 3 >= 3, select it.

3. Final Answer:

   1 -> Option C

4. Quick Check:

   One removal needed to avoid overlap [OK]

Hint: Pick intervals ending earliest, remove overlaps [OK]

Common Mistakes:

• Removing too many intervals
• Not sorting intervals before processing
• Confusing overlapping intervals with touching intervals