# 436. Find Right Interval

Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the “right” of i.

For any interval i, you need to store the minimum interval j’s index, which means that the interval j has the minimum start point to build the “right” relationship for interval i. If the interval j doesn’t exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array.

Note:

1. You may assume the interval’s end point is always bigger than its start point.
2. You may assume none of these intervals have the same start point.

Example 1:

``````Input: [ [1,2] ]

Output: [-1]

Explanation: There is only one interval in the collection, so it outputs -1.
``````

Example 2:

``````Input: [ [3,4], [2,3], [1,2] ]

Output: [-1, 0, 1]

Explanation: There is no satisfied "right" interval for [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point;
For [1,2], the interval [2,3] has minimum-"right" start point.
``````

Example 3:

``````Input: [ [1,4], [2,3], [3,4] ]

Output: [-1, 2, -1]

Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point.
``````

``````class Solution {
public:
vector<int> findRightInterval(vector<vector<int>>& intervals) {
vector<int> res, starts;
unordered_map<int, int> m;
for (int i = 0; i < intervals.size(); ++i) {
m[intervals[i][0]] = i;
starts.push_back(intervals[i][0]);
}
sort(starts.rbegin(), starts.rend());
for (auto interval : intervals) {
int i = 0;
for (; i < starts.size(); ++i) {
if (starts[i] < interval[1]) break;
}
res.push_back((i > 0) ? m[starts[i - 1]] : -1);
}
return res;
}
};
``````

``````class Solution {
public:
vector<int> findRightInterval(vector<vector<int>>& intervals) {
vector<int> res;
map<int, int> m;
for (int i = 0; i < intervals.size(); ++i) {
m[intervals[i][0]] = i;
}
for (auto interval : intervals) {
auto it = m.lower_bound(interval[1]);
if (it == m.end()) res.push_back(-1);
else res.push_back(it->second);
}
return res;
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/436

Non-overlapping Intervals

Insert Interval

Merge Intervals

https://leetcode.com/problems/find-right-interval/

https://leetcode.com/problems/find-right-interval/discuss/91819/C%2B%2B-map-solution

https://leetcode.com/problems/find-right-interval/discuss/91789/Java-clear-O(n-logn)-solution-based-on-TreeMap

LeetCode All in One 题目讲解汇总(持续更新中…)

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