Non Overlapping Intervals Minimum Removal in DSA C++ - Build from Scratch

Non Overlapping Intervals Minimum Removal

📖 Scenario: You are organizing a conference with multiple talks scheduled in different time intervals. Some talks overlap, and you want to remove the minimum number of talks so that no two talks overlap.

🎯 Goal: Build a program that finds the minimum number of intervals to remove to make the rest non-overlapping.

📋 What You'll Learn

Create a vector of pairs representing intervals with exact values

Add a variable to count removals

Implement sorting of intervals by their end time

Use a loop to find minimum removals based on interval overlaps

Print the minimum number of intervals to remove

💡 Why This Matters

🌍 Real World

Scheduling talks, meetings, or tasks without conflicts is common in event planning and calendar apps.

💼 Career

Understanding interval scheduling and greedy algorithms is important for software engineers working on optimization and resource allocation problems.

1

Create the intervals vector

Create a vector of pairs called intervals with these exact intervals: {1, 3}, {2, 4}, {3, 5}, {6, 7}, {8, 10}, {9, 11}.

DSA C++

std::vector> intervals = {
    // Your code here
};

Hint

Use std::vector> and initialize with the given intervals.

2

Add a removal counter variable

Add an integer variable called removals and set it to 0.

DSA C++

std::vector> intervals = {
    {1, 3}, {2, 4}, {3, 5}, {6, 7}, {8, 10}, {9, 11}
};

// Your code here

Hint

Declare int removals = 0; to count how many intervals you remove.

3

Sort intervals and find minimum removals

Sort intervals by their end time using std::sort with a lambda. Then create an integer prev_end set to the end of the first interval. Use a for loop starting from index 1 to check if current interval start is less than prev_end. If yes, increment removals. Otherwise, update prev_end to current interval's end.

DSA C++

std::vector> intervals = {
    {1, 3}, {2, 4}, {3, 5}, {6, 7}, {8, 10}, {9, 11}
};

int removals = 0;

// Your code here

Hint

Sort by end time, then loop from second interval to check overlaps and count removals.

4

Print the minimum number of removals

Print the value of removals using std::cout.

DSA C++

std::vector> intervals = {
    {1, 3}, {2, 4}, {3, 5}, {6, 7}, {8, 10}, {9, 11}
};

int removals = 0;

std::sort(intervals.begin(), intervals.end(), [](const std::pair& a, const std::pair& b) {
    return a.second < b.second;
});

int prev_end = intervals[0].second;
for (int i = 1; i < intervals.size(); i++) {
    if (intervals[i].first < prev_end) {
        removals++;
    } else {
        prev_end = intervals[i].second;
    }
}

// Your code here

Hint

Use std::cout << removals << std::endl; to print the result.

Practice

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

easy

A. To merge all intervals into one

B. To find the maximum number of overlapping intervals

C. To remove the minimum number of intervals so that the remaining intervals do not overlap

D. To sort intervals by their start time

Solution

1. Step 1: Understand the problem goal

   The problem asks to remove as few intervals as possible so that no two intervals overlap.

2. Step 2: Clarify what non-overlapping means

   Non-overlapping means for any two intervals, one ends before the other starts.

3. Final Answer:

   To remove the minimum number of intervals so that the remaining intervals do not overlap -> Option C

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
• Sorting intervals without removing

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

easy

A. std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.end < b.end; });

B. std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.start < b.start; });

C. std::sort(intervals.begin(), intervals.end());

D. std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.end > b.end; });

Solution

1. Step 1: Identify sorting criteria for minimum removal

   We sort intervals by their end time ascending to greedily pick intervals that finish earliest.

2. Step 2: Check code correctness

   std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.end < b.end; }); uses a lambda comparing a.end < b.end, which is correct for sorting by end time ascending.

3. Final Answer:

   std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.end < b.end; }); -> Option A

4. Quick Check:

   Sort by end time ascending = std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.end < b.end; }); [OK]

Hint: Sort intervals by end time ascending for greedy approach [OK]

Common Mistakes:

• Sorting by start time instead of end time
• Sorting in descending order of end time
• Using default sort without comparator

3. What is the output of the following C++ code snippet?

struct Interval { int start, end; };
std::vector<Interval> intervals = {{1,3}, {2,4}, {3,5}};
std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.end < b.end; });
int count = 0, end_time = INT_MIN;
for (auto &interval : intervals) {
  if (interval.start >= end_time) {
    end_time = interval.end;
  } else {
    count++;
  }
}
std::cout << count << std::endl;

medium

A. 2

B. 1

C. 0

D. 3

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals by end: (1,3), (2,4), (3,5)

2. Step 2: Iterate and count removals

   Start with end_time = -∞ - Interval (1,3): start 1 >= -∞, keep, end_time=3 - Interval (2,4): start 2 < 3, overlap, count=1 - Interval (3,5): start 3 >= 3, keep, end_time=5

3. Final Answer:

   1 -> Option B

4. Quick Check:

   One overlap removed = 1 [OK]

Hint: Count overlaps by comparing start with last end time [OK]

Common Mistakes:

• Not updating end_time correctly
• Counting non-overlapping intervals as removals
• Sorting incorrectly before iteration

4. Identify the bug in the following C++ code for minimum interval removal:

std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.start < b.start; });
int count = 0, end_time = INT_MIN;
for (auto &interval : intervals) {
  if (interval.start >= end_time) {
    end_time = interval.end;
  } else {
    count++;
  }
}
return count;

medium

A. Not initializing count to 1 causes off-by-one error

B. Using INT_MIN for end_time causes overflow

C. Missing check for empty intervals vector

D. Sorting by start time instead of end time causes incorrect removal count

Solution

1. Step 1: Check sorting criteria

   The code sorts intervals by start time, but greedy approach requires sorting by end time to minimize removals.

2. Step 2: Understand impact on counting

   Sorting by start time can cause counting more removals than needed because intervals ending earlier are not prioritized.

3. Final Answer:

   Sorting by start time instead of end time causes incorrect removal count -> Option D

4. Quick Check:

   Sort by end time, not start time [OK]

Hint: Always sort by end time for minimum removal [OK]

Common Mistakes:

• Sorting by start time instead of end time
• Assuming INT_MIN causes overflow here
• Ignoring empty input checks

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. 1

B. 2

C. 3

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

   Keep [1,2] (end=2) - [2,3]: start 2 >= 2, keep - [1,3]: start 1 < 2 (overlaps with [1,2]), remove (count=1) - [3,4]: start 3 >= 3, keep Total removals so far: 1 But [1,3] overlaps with [1,2], removing it once is enough. Check if any other removals needed: none Total removals = 1

3. Step 3: Re-examine intervals order

   Actually, sorting by end time ascending is: [1,2], [2,3], [1,3], [3,4] But [1,3] overlaps with [1,2], so removing it once suffices. If we keep [1,2], [2,3], [3,4], we remove [1,3] only once. So minimum removals = 1

4. Final Answer:

   1 -> Option A

5. Quick Check:

   Remove overlapping [1,3] only once [OK]

Hint: Sort by end time, greedily pick intervals, count removals [OK]

Common Mistakes:

• Removing too many intervals unnecessarily
• Not sorting by end time
• Confusing overlapping intervals