# Find Right Interval Problem

## Description

LeetCode Problem 436.

You are given an array of intervals, where intervals[i] = [start_i, end_i] and each start_i is unique.

The right interval for an interval i is an interval j such that start_j >= end_i and start_j is minimized.

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:

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Input: intervals = [[1,2]]
Output: [-1]
Explanation: There is only one interval in the collection, so it outputs -1.
``````

Example 2:

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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 start_0= 3 is the smallest start that is >= end_1= 3.
The right interval for [1,2] is [2,3] since start_1= 2 is the smallest start that is >= end_2= 2.
``````

Example 3:

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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 start_2 = 3 is the smallest start that is >= end_1= 3.
``````

Constraints:

• 1 <=intervals.length <= 2 * 10^4
• intervals[i].length == 2
• -10^6 <= start_i <= end_i <= 10^6
• The start pointof each interval is unique.

## Sample C++ Code

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