Non Overlapping Intervals Minimum Removal in DSA C - Time & Space Complexity

Time Complexity: Non Overlapping Intervals Minimum Removal

O(n log n)

Understanding Time Complexity

We want to find how fast the algorithm runs when removing intervals to avoid overlaps.

How does the time needed grow as the number of intervals increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Sort intervals by end time
qsort(intervals, n, sizeof(Interval), compareEnd);

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

This code sorts intervals by their end time, then counts how many intervals must be removed to avoid overlaps.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting the intervals array.
• How many times: Sorting compares elements multiple times, then a single loop runs through all intervals once.

How Execution Grows With Input

As the number of intervals grows, sorting takes more time, and the loop checks each interval once.

Input Size (n)	Approx. Operations
10	About 30 to 40 operations
100	About 700 to 800 operations
1000	About 10,000 to 12,000 operations

Pattern observation: Operations grow faster than the input size but not as fast as its square; sorting dominates growth.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a bit faster than the number of intervals because sorting takes more work as input grows.

Common Mistake

[X] Wrong: "The loop checking overlaps is the slowest part, so complexity is O(n)."

[OK] Correct: Sorting the intervals takes more time than the single loop, so it controls the overall speed.

Interview Connect

Understanding this helps you explain how sorting affects performance and why greedy choices after sorting are efficient.

Self-Check

"What if the intervals were already sorted by start time? How would the time complexity change?"

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