How do you solve Find Right Interval on LeetCode?

Find Right Interval (LeetCode #436) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting intervals helps in efficiently finding the right interval using binary search.

What is the brute force solution for Find Right Interval?

The brute force approach for Find Right Interval is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each interval against every other interval to find the right interval. This is straightforward but inefficient, especially for larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Find Right Interval?

Find Right Interval uses the following concepts: Array, Binary Search, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Right Interval?

Always clarify the problem constraints and edge cases before diving into coding. Think about the efficiency of your solution and be ready to explain your thought process. Practice explaining your code as you write it, as this helps in interviews.

What are common mistakes when solving Find Right Interval?

Not considering the case where no right interval exists, leading to incorrect results. Confusing the start and end times when comparing intervals.

#436

Find Right Interval

Medium

ArrayBinary SearchSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

By sorting the intervals based on their start times and using binary search, we can efficiently find the right interval for each interval in O(n log n) time.

⚙️

Algorithm

5 steps

1Step 1: Create a list of tuples containing each interval's start and its original index.
2Step 2: Sort this list based on the start times of the intervals.
3Step 3: For each interval, use binary search to find the smallest start time that is greater than or equal to the end time of the current interval.
4Step 4: Store the corresponding original index in the result array.
5Step 5: Return the result array.

solution.py16 lines

1def findRightInterval(intervals):
2    indexed_intervals = [(start, i) for i, (start, end) in enumerate(intervals)]
3    indexed_intervals.sort()
4    result = [-1] * len(intervals)
5    for start, i in indexed_intervals:
6        end = intervals[i][1]
7        left, right = 0, len(intervals)
8        while left < right:
9            mid = (left + right) // 2
10            if intervals[mid][0] < end:
11                left = mid + 1
12            else:
13                right = mid
14        if left < len(intervals):
15            result[i] = indexed_intervals[left][1]
16    return result

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the binary search for each interval takes O(log n), making it efficient for larger datasets. The space complexity is O(n) because we store the indexed intervals.

1Sorting intervals helps in efficiently finding the right interval using binary search.
2Unique start times allow for a straightforward mapping of intervals to their indices.

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