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 we can decide the minimum number of intervals to remove so none overlap.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <vector>
#include <algorithm>

int eraseOverlapIntervals(std::vector<std::pair<int,int>>& intervals) {
    if (intervals.empty()) return 0;
    std::sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) { return a.second < b.second; });
    int count = 0, end = intervals[0].second;
    for (int i = 1; i < intervals.size(); ++i) {
        if (intervals[i].first < end) {
            ++count;
        } else {
            end = intervals[i].second;
        }
    }
    return count;
}
    

This code sorts intervals by their end time, then counts how many overlap to remove the minimum.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting the intervals by their end time.
• How many times: Sorting takes about n log n steps for n intervals.
• Secondary operation: One loop through all intervals to count overlaps.
• How many times: This loop runs n-1 times, once per interval after the first.

How Execution Grows With Input

Sorting grows faster than the loop, so it dominates the time.

Input Size (n)	Approx. Operations
10	About 10 * log 10 = 33 steps
100	About 100 * log 100 = 664 steps
1000	About 1000 * log 1000 = 9966 steps

Pattern observation: As n grows, sorting steps grow a bit faster than just n, roughly n times log n.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a bit faster than the number of intervals, mainly due to sorting.

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 loop, so it controls the overall speed.

Interview Connect

Understanding how sorting affects time helps you explain your solution clearly and shows you know what part costs the most time.

Self-Check

"What if the intervals were already sorted by end 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 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