Why Intervals Are a Common Problem Pattern in DSA C - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a struct for an interval with start and end.

DSA C

typedef struct {
    int start;
    int [1];
} Interval;

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

Avalue

Blength

Cend

Drange

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'length' instead of 'end' for the second member.

The struct for an interval typically has 'start' and 'end' to mark boundaries.

2fill in blank

medium

Complete the code to check if two intervals overlap.

DSA C

int isOverlap(Interval a, Interval b) {
    return !(a.end <= b.[1] || b.end <= a.start);
}

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

Avalue

Bstart

Clength

Drange

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with 'length' or 'value' which are not interval boundaries.

To check overlap, compare the end of one with the start of the other.

3fill in blank

hard

Fix the error in the function that merges two overlapping intervals.

DSA C

Interval mergeIntervals(Interval a, Interval b) {
    Interval merged;
    merged.start = a.start < b.start ? a.start : b.[1];
    merged.end = a.end > b.end ? a.end : b.end;
    return merged;
}

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

Astart

Brange

Cvalue

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'end' or 'length' instead of 'start' for the minimum start.

When merging, the start is the minimum of both starts.

4fill in blank

hard

Fill both blanks to create a function that sorts intervals by their start time.

DSA C

int compareIntervals(const void *a, const void *b) {
    Interval *i1 = (Interval *)[1];
    Interval *i2 = (Interval *)[2];
    return i1->start - i2->start;
}

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

Aa

Bb

Cstart

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'start' or 'end' as variable names instead of 'a' and 'b'.

In qsort compare function, cast the void pointers to Interval pointers using 'a' and 'b'.

5fill in blank

hard

Fill all three blanks to create a function that merges a list of intervals.

DSA C

int mergeAllIntervals(Interval intervals[], int n, Interval result[]) {
    if (n == 0) return 0;
    qsort(intervals, n, sizeof(Interval), compareIntervals);
    int idx = 0;
    result[idx] = intervals[0];
    for (int i = 1; i < n; i++) {
        if (result[idx].end >= intervals[i].[1]) {
            if (result[idx].end < intervals[i].[2]) {
                result[idx].end = intervals[i].[3];
            }
        } else {
            idx++;
            result[idx] = intervals[i];
        }
    }
    return idx + 1;
}

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

Astart

Bend

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up start and end in comparisons and assignments.

The function merges intervals by checking if current interval's start is within the last merged interval's end, then updates the end if needed.

Practice

1. Why are intervals commonly used in problems involving scheduling or resource allocation?

easy

A. Because intervals only contain single points, making them easy to handle.

B. Because intervals are always sorted by default in data structures.

C. Because intervals do not overlap in any problem scenario.

D. Because intervals represent time ranges or continuous segments that need to be managed or optimized.

Solution

1. Step 1: Understand what intervals represent

   Intervals represent continuous ranges, such as time slots or segments, which are common in scheduling and resource allocation.

2. Step 2: Recognize why intervals are useful

   Managing these continuous ranges helps optimize usage, avoid conflicts, and allocate resources efficiently.

3. Final Answer:

   Because intervals represent time ranges or continuous segments that need to be managed or optimized. -> Option D

4. Quick Check:

   Intervals = continuous ranges [OK]

Hint: Intervals model continuous ranges like time slots [OK]

Common Mistakes:

• Thinking intervals are always sorted by default
• Assuming intervals contain only single points
• Believing intervals never overlap

2. Which of the following is the correct syntax to declare an interval struct in C with start and end integers?

easy

A. struct Interval { int start; int end }

B. struct Interval { int start, end };

C. struct Interval { int start; int end; };

D. struct Interval (int start; int end;);

Solution

1. Step 1: Recall C struct syntax

   A struct declaration requires braces {}, semicolons after each member, and ends with a semicolon.

2. Step 2: Check each option

   struct Interval { int start; int end; }; correctly uses braces, semicolons after members, and ends with a semicolon. struct Interval { int start, end }; misses semicolon after members. struct Interval { int start; int end } misses semicolon after closing brace. struct Interval (int start; int end;); uses parentheses incorrectly.

3. Final Answer:

   struct Interval { int start; int end; }; -> Option C

4. Quick Check:

   Struct syntax = braces + semicolons [OK]

Hint: Struct members end with semicolons; struct ends with semicolon [OK]

Common Mistakes:

• Omitting semicolons after members
• Forgetting semicolon after struct closing brace
• Using parentheses instead of braces

3. What is the output of the following C code that merges two intervals if they overlap?

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

Interval mergeIntervals(Interval a, Interval b) {
    if (a.end < b.start) return a;
    if (b.end < a.start) return b;
    Interval merged = {a.start < b.start ? a.start : b.start, a.end > b.end ? a.end : b.end};
    return merged;
}

int main() {
    Interval i1 = {1, 5};
    Interval i2 = {4, 8};
    Interval result = mergeIntervals(i1, i2);
    printf("%d %d\n", result.start, result.end);
    return 0;
}

medium

A. 1 8

B. 4 5

C. 1 5

D. 4 8

Solution

1. Step 1: Check if intervals overlap

   i1 = [1,5], i2 = [4,8]. Since 5 >= 4, they overlap.

2. Step 2: Calculate merged interval

   Start = min(1,4) = 1, End = max(5,8) = 8.

3. Final Answer:

   1 8 -> Option A

4. Quick Check:

   Overlap merge = [1,8] [OK]

Hint: Merge intervals by min start and max end if overlapping [OK]

Common Mistakes:

• Returning the first interval without merging
• Using max start or min end instead of min start and max end
• Ignoring overlap condition

4. Identify the bug in the following C function that checks if two intervals overlap:

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

int isOverlap(Interval a, Interval b) {
    if (a.end < b.start || b.end < a.start) {
        return 0;
    }
    return 1;
}

medium

A. The function should use '<' instead of '<=' to consider touching intervals as non-overlapping.

B. The function incorrectly treats touching intervals as overlapping.

C. The function misses the case when intervals are equal.

D. The function should return 2 for overlapping intervals.

Solution

1. Step 1: Understand overlap condition

   Intervals overlap if they share interior points. Touching intervals (a.end == b.start) usually do NOT overlap in scheduling problems.

2. Step 2: Analyze the condition

   The code uses '<' which treats touching intervals (a.end == b.start) as overlapping (returns 1), since a.end < b.start is false. This is usually incorrect; should use '<=' for no-overlap condition.

3. Final Answer:

   The function incorrectly treats touching intervals as overlapping. -> Option B

4. Quick Check:

   Touching = non-overlap: use <= [OK]

Hint: Use '<=' to exclude touching intervals from overlap [OK]

Common Mistakes:

• Confusing touching intervals as overlapping
• Returning wrong values for overlap
• Ignoring equal intervals case

5. Given a list of intervals, how would you efficiently find the minimum number of meeting rooms required to hold all meetings without overlap?

hard

A. Sort intervals by start time, then use a min-heap to track earliest end times and count rooms needed.

B. Sort intervals by end time and assign rooms in order without tracking overlaps.

C. Count the total number of intervals and use that as the number of rooms.

D. Merge all intervals into one and assign a single room.

Solution

1. Step 1: Sort intervals by start time

   Sorting helps process meetings in chronological order.

2. Step 2: Use a min-heap to track earliest end times

   Min-heap stores end times of ongoing meetings; if a meeting starts after the earliest end, reuse room; else add a new room.

3. Step 3: Count rooms needed

   The size of the min-heap at any time is the number of rooms needed.

4. Final Answer:

   Sort intervals by start time, then use a min-heap to track earliest end times and count rooms needed. -> Option A

5. Quick Check:

   Min-heap tracks rooms efficiently [OK]

Hint: Use min-heap on end times after sorting by start [OK]

Common Mistakes:

• Ignoring overlap and just counting intervals
• Merging intervals loses overlap info
• Sorting by end time alone is insufficient