Introduction

In this blog post, we will explore how to solve the LeetCode problem "436" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an array of intervals, where intervals[i] = [starti, endi] and each starti is unique. The right interval for an interval i is an interval j such that startj >= endi and startj is minimized. Note that i may equal j. Return an array of right interval indices for each interval i. If no right interval exists for interval i, then put -1 at index i. Example 1: Input: intervals = [[1,2]] Output: [-1] Explanation: There is only one interval in the collection, so it outputs -1. Example 2: Input: intervals = [[3,4],[2,3],[1,2]] Output: [-1,0,1] Explanation: There is no right interval for [3,4]. The right interval for [2,3] is [3,4] since start0 = 3 is the smallest start that is >= end1 = 3. The right interval for [1,2] is [2,3] since start1 = 2 is the smallest start that is >= end2 = 2. Example 3: Input: intervals = [[1,4],[2,3],[3,4]] Output: [-1,2,-1] Explanation: There is no right interval for [1,4] and [3,4]. The right interval for [2,3] is [3,4] since start2 = 3 is the smallest start that is >= end1 = 3. Constraints: 1 <= intervals.length <= 2 * 104 intervals[i].length == 2 -106 <= starti <= endi <= 106 The start point of each interval is unique.

Explanation

Here's the solution to the problem, broken down into key points, complexity analysis, and the complete Python code.

• Core Idea: The most efficient approach involves sorting the intervals based on their start times, then using binary search to find the right interval for each original interval based on its end time.
• Index Preservation: Store the original index of each interval before sorting so that we can place the result in the correct position in the output array.

• Binary Search Application: For each interval, use binary search on the sorted start times to find the smallest start time that is greater than or equal to the interval's end time.

• Complexity: O(N log N) time, O(N) space, where N is the number of intervals.

Code

    def findRightInterval(intervals):
    """
    Finds the right interval for each interval in a given array of intervals.

    Args:
        intervals: A list of intervals, where each interval is a list of two integers [start, end].

    Returns:
        A list of integers, where the i-th element is the index of the right interval for the i-th interval.
        If no right interval exists, the element is -1.
    """

    n = len(intervals)
    indexed_intervals = sorted([(intervals[i][0], i) for i in range(n)])  # Sort by start time, keep original index
    result = [-1] * n

    for i in range(n):
        end_i = intervals[i][1]
        l, r = 0, n - 1
        right_index = -1

        while l <= r:
            mid = (l + r) // 2
            if indexed_intervals[mid][0] >= end_i:
                right_index = indexed_intervals[mid][1]
                r = mid - 1  # Try to find an even smaller start time
            else:
                l = mid + 1

        result[i] = right_index

    return result