How do you solve Count Integers in Intervals on LeetCode?

Count Integers in Intervals (LeetCode #2276) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Merging overlapping intervals is crucial for efficiency.

What is the brute force solution for Count Integers in Intervals?

The brute force approach for Count Integers in Intervals is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves maintaining a list of all integers covered by the intervals. Every time we add a new interval, we simply add all integers in that range to our list and count the unique integers. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Integers in Intervals?

Count Integers in Intervals uses the following concepts: Design, Segment Tree, Ordered Set. The recommended patterns to study are: Interval Merging, Dynamic Programming.

What are the interview tips for Count Integers in Intervals?

Explain your thought process clearly. Discuss the trade-offs of different approaches. Practice merging intervals with different scenarios.

What are common mistakes when solving Count Integers in Intervals?

Failing to merge intervals correctly. Not considering edge cases like fully overlapping intervals.

#2276

Count Integers in Intervals

Hard

DesignSegment TreeOrdered SetInterval MergingDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a list to store intervals and merges overlapping intervals whenever a new interval is added. This way, we maintain a compact representation of the intervals, allowing us to count the total integers efficiently.

⚙️

Algorithm

3 steps

1Step 1: Maintain a list of non-overlapping intervals.
2Step 2: When adding a new interval, merge it with any overlapping intervals.
3Step 3: After merging, calculate the total count of integers covered by the intervals.

solution.py23 lines

1class CountIntervals:
2    def __init__(self):
3        self.intervals = []
4
5    def add(self, left: int, right: int) -> None:
6        new_intervals = []
7        i = 0
8        while i < len(self.intervals) and self.intervals[i][1] < left:
9            new_intervals.append(self.intervals[i])
10            i += 1
11        while i < len(self.intervals) and self.intervals[i][0] <= right:
12            left = min(left, self.intervals[i][0])
13            right = max(right, self.intervals[i][1])
14            i += 1
15        new_intervals.append((left, right))
16        new_intervals.extend(self.intervals[i:])
17        self.intervals = new_intervals
18
19    def count(self) -> int:
20        total = 0
21        for left, right in self.intervals:
22            total += right - left + 1
23        return total

ℹ

Complexity note: The time complexity is O(n) for adding intervals since we may need to merge existing intervals. The space complexity is O(n) because we store the merged intervals.

1Merging overlapping intervals is crucial for efficiency.
2Using a list to maintain non-overlapping intervals simplifies counting.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.