LeetCode 436: Find Right Interval Solution

Master LeetCode problem 436 (Find Right Interval), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

436. Find Right Interval

Medium

Array Binary Search Sorting

Problem Explanation

Explanation:

To solve this problem, we can first sort the intervals based on their start points. Then, for each interval, we can perform a binary search to find the right interval by looking for the smallest start point that is greater than or equal to the current interval's end point.

Sort the intervals based on their start points.
For each interval, perform a binary search on the sorted intervals to find the right interval.
If a right interval is found, store its index in the result array. Otherwise, store -1.

Time complexity: O(n log n) where n is the number of intervals (due to sorting and binary searching) Space complexity: O(n) for storing the result array

Solution Code

class Solution {
    public int[] findRightInterval(int[][] intervals) {
        TreeMap<Integer, Integer> map = new TreeMap<>();
        int n = intervals.length;
        int[] result = new int[n];
        
        for (int i = 0; i < n; i++) {
            map.put(intervals[i][0], i);
        }
        
        for (int i = 0; i < n; i++) {
            Map.Entry<Integer, Integer> entry = map.ceilingEntry(intervals[i][1]);
            result[i] = (entry != null) ? entry.getValue() : -1;
        }
        
        return result;
    }
}

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