Find Right Interval – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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.

C++ Find Right Interval LeetCode Solution

class Solution {
public:
    vector<int> findRightInterval(vector<vector<int>>& intervals) {
        vector<int> ans(intervals.size());
        map<int, int> m;
        for(int i = 0; i < intervals.size(); i++)
            m[intervals[i][0]] = i;
        for(int i = 0; i < intervals.size(); i++) 
               ans[i]  = m.lower_bound(intervals[i][1])!=end(m)?m.lower_bound(intervals[i][1]) -> second:-1;
        return ans;
    }
};

Java Find Right Interval LeetCode Solution

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

Python 3 Find Right Interval LeetCode Solution

def findRightInterval(self, intervals):
    l = sorted((e.start, i) for i, e in enumerate(intervals))
    res = []
    for e in intervals:
        r = bisect.bisect_left(l, (e.end,))
        res.append(l[r][1] if r < len(l) else -1)
    return res

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