Non Overlapping Intervals Minimum Removal in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort intervals by their end time.

DSA C

int compare(const void *a, const void *b) {
    int *intervalA = *(int **)a;
    int *intervalB = *(int **)b;
    return intervalA[[1]] - intervalB[[1]];
}

Drag options to blanks, or click blank then click option'

A-1

B0

C1

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by start time instead of end time.

Using wrong index for end time.

We sort intervals by their end time, which is at index 1 in each interval array.

2fill in blank

medium

Complete the code to initialize the count of non-overlapping intervals.

DSA C

int count = 1;
int prev_end = intervals[0][[1]];

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A1

B2

C0

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using start time instead of end time.

Initializing count to 0 instead of 1.

We start with the first interval's end time at index 1 to compare with others.

3fill in blank

hard

Fix the error in the condition to check if intervals overlap.

DSA C

for (int i = 1; i < intervalsSize; i++) {
    if (intervals[i][0] >= [1]) {
        count++;
        prev_end = intervals[i][1];
    }
}

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Aintervals[i][1]

Bcount

Cintervals[i][0]

Dprev_end

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing start time with start time.

Using wrong variable in condition.

We compare the current interval's start time with the previous interval's end time to check overlap.

4fill in blank

hard

Fill both blanks to calculate the minimum number of intervals to remove.

DSA C

int minRemoval = [1] - [2];
return minRemoval;

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AintervalsSize

Bcount

Cprev_end

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Subtracting wrong variables.

Returning count instead of minimum removals.

Minimum removals = total intervals - count of non-overlapping intervals.

5fill in blank

hard

Fill all three blanks to complete the function for minimum removals.

DSA C

int eraseOverlapIntervals(int** intervals, int intervalsSize, int* intervalsColSize) {
    qsort(intervals, intervalsSize, sizeof(int*), [1]);
    int count = 1;
    int prev_end = intervals[0][[2]];
    for (int i = 1; i < intervalsSize; i++) {
        if (intervals[i][0] >= [3]) {
            count++;
            prev_end = intervals[i][1];
        }
    }
    return intervalsSize - count;
}

Drag options to blanks, or click blank then click option'

Acompare

B1

Cprev_end

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong function name in qsort.

Using wrong indices for start/end times.

Incorrect overlap condition.

We use compare function for sorting, index 1 for end time, and prev_end to check overlap.

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