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 an interval struct with start and end.

DSA C++

struct Interval {
    int [1];
    int end;
};

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

Astart

Bbegin

Cleft

Drange

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'begin' or 'left' instead of 'start' which is the conventional name.

The interval struct commonly uses 'start' to represent the beginning of the interval.

2fill in blank

medium

Complete the code to sort intervals by their start time.

DSA C++

bool compareIntervals(const Interval &a, const Interval &b) {
    return a.[1] < b.start;
}

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Aend

Blength

Cstart

Dduration

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing by 'end' instead of 'start' which changes the sorting order.

Sorting intervals by their 'start' time helps in many interval problems.

3fill in blank

hard

Fix the error in the code to merge overlapping intervals.

DSA C++

vector<Interval> mergeIntervals(vector<Interval> &intervals) {
    if (intervals.empty()) return {};
    sort(intervals.begin(), intervals.end(), compareIntervals);
    vector<Interval> merged;
    merged.push_back(intervals[0]);
    for (int i = 1; i < intervals.size(); i++) {
        if (merged.back().end >= intervals[i].[1]) {
            merged.back().end = max(merged.back().end, intervals[i].end);
        } else {
            merged.push_back(intervals[i]);
        }
    }
    return merged;
}

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Aduration

Bend

Clength

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing end with end instead of end with start causes wrong merging.

To check overlap, compare the end of last merged interval with the start of current interval.

4fill in blank

hard

Fill both blanks to create a map of intervals keyed by their start time and storing their end time.

DSA C++

map<int, int> intervalMap = {
    {intervals[0].[1], intervals[0].[2]
};

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Astart

Bend

Clength

Dduration

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'length' or 'duration' which are not fields in the struct.

The map keys intervals by their start and stores the end as value.

5fill in blank

hard

Fill all three blanks to create a vector of intervals from two separate vectors of starts and ends.

DSA C++

vector<Interval> createIntervals(const vector<int> &starts, const vector<int> &ends) {
    vector<Interval> intervals;
    for (int i = 0; i < starts.size(); i++) {
        intervals.push_back([1], [2]);
    }
    sort(intervals.begin(), intervals.end(), [3]);
    return intervals;
}

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Astarts[i]

Bends[i]

CcompareIntervals

Dintervals[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using intervals[i] inside push_back or wrong comparator in sort.

We use starts[i] and ends[i] to create intervals and sort them using compareIntervals.

Practice

1. Why are interval problems common in algorithms and data structures?

easy

A. Because many real-world problems involve ranges or periods of time.

B. Because intervals are easier to sort than other data types.

C. Because intervals always have fixed lengths.

D. Because intervals do not overlap in any problem.

Solution

1. Step 1: Understand the nature of intervals

   Intervals represent ranges, such as time periods or numeric ranges, which appear in many real-world scenarios like scheduling or booking.

2. Step 2: Recognize why intervals are common

   Since many problems require handling overlapping or non-overlapping ranges, intervals become a natural pattern to solve these efficiently.

3. Final Answer:

   Because many real-world problems involve ranges or periods of time. -> Option A

4. Quick Check:

   Intervals model ranges = Because many real-world problems involve ranges or periods of time. [OK]

Hint: Think about real-life ranges like meetings or bookings [OK]

Common Mistakes:

• Assuming intervals always have fixed lengths
• Thinking intervals never overlap
• Believing intervals are only about sorting

2. Which of the following is the correct C++ syntax to represent an interval as a pair of integers start and end?

easy

A. std::pair interval = {start, end};

B. std::vector<int> interval = {start, end};

C. int interval = start, end;

D. std::map interval = {start, end};

Solution

1. Step 1: Identify the correct container for two related integers

   A pair holds exactly two values together, suitable for start and end.

2. Step 2: Check syntax correctness

   std::pair interval = {start, end}; correctly initializes the pair with start and end.

3. Final Answer:

   std::pair interval = {start, end}; -> Option A

4. Quick Check:

   Pair holds two ints = std::pair interval = {start, end}; [OK]

Hint: Use std::pair for two related values like interval ends [OK]

Common Mistakes:

• Using vector which can hold many elements, not just two
• Declaring two ints separated by comma incorrectly
• Using map which is for key-value pairs, not intervals

3. What is the output of the following C++ code that merges overlapping intervals?

std::vector> intervals = {{1,3}, {2,6}, {8,10}, {15,18}};
std::sort(intervals.begin(), intervals.end());
std::vector> merged;
for (auto &interval : intervals) {
  if (merged.empty() || merged.back().second < interval.first) {
    merged.push_back(interval);
  } else {
    merged.back().second = std::max(merged.back().second, interval.second);
  }
}
for (auto &i : merged) {
  std::cout << "{" << i.first << "," << i.second << "} ";
}

medium

A. {1,10} {15,18}

B. {1,3} {2,6} {8,10} {15,18}

C. {1,6} {8,10} {15,18}

D. {1,6} {2,10} {15,18}

Solution

1. Step 1: Sort intervals by start time

   Intervals sorted: {1,3}, {2,6}, {8,10}, {15,18}

2. Step 2: Merge overlapping intervals

   {1,3} and {2,6} overlap, merged to {1,6}. Others remain separate.

3. Final Answer:

   {1,6} {8,10} {15,18} -> Option C

4. Quick Check:

   Merged intervals = {1,6} {8,10} {15,18} [OK]

Hint: Sort intervals first, then merge overlapping ones [OK]

Common Mistakes:

• Not sorting intervals before merging
• Incorrectly merging non-overlapping intervals
• Printing intervals without merging

4. The following code is intended to merge overlapping intervals but has a bug. What is the bug?

std::vector> intervals = {{1,4}, {4,5}};
std::sort(intervals.begin(), intervals.end());
std::vector> merged;
for (auto &interval : intervals) {
  if (merged.empty() || merged.back().second <= interval.first) {
    merged.push_back(interval);
  } else {
    merged.back().second = std::max(merged.back().second, interval.second);
  }
}

medium

A. The max function is used incorrectly.

B. The condition should use '<' instead of '<=' to merge touching intervals.

C. The vector intervals is not sorted before merging.

D. The merged vector is not initialized.

Solution

1. Step 1: Analyze the merging condition

   The condition uses merged.back().second <= interval.first, which treats touching intervals as non-overlapping.

2. Step 2: Correct the condition for merging touching intervals

   Using '<' instead of '<=' merges intervals that touch at endpoints, which is usually desired.

3. Final Answer:

   The condition should use '<' instead of '<=' to merge touching intervals. -> Option B

4. Quick Check:

   Use '<' to merge touching intervals = The condition should use '<' instead of '<=' to merge touching intervals. [OK]

Hint: Use '<' not '<=' to merge intervals that just touch [OK]

Common Mistakes:

• Ignoring touching intervals as overlapping
• Not sorting intervals before merging
• Misusing max function

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

hard

A. Sort intervals by end time, merge all intervals, count merged intervals.

B. Count intervals with the longest duration only.

C. Use a stack to push intervals and pop when they overlap.

D. Sort intervals by start time, use a min-heap to track end times, count max simultaneous meetings.

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 ending meeting

   Push end times into a min-heap; if a meeting starts after the earliest end, reuse room; else add room.

3. Step 3: Count max simultaneous meetings

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

4. Final Answer:

   Sort intervals by start time, use a min-heap to track end times, count max simultaneous meetings. -> Option D

5. Quick Check:

   Min rooms = max overlaps tracked by min-heap = Sort intervals by start time, use a min-heap to track end times, count max simultaneous meetings. [OK]

Hint: Track earliest finishing meeting with min-heap to count rooms [OK]

Common Mistakes:

• Merging intervals instead of counting overlaps
• Using stack which doesn't track end times properly
• Ignoring sorting by start time