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 * 104intervals[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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