Find Right Interval

Description

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.

Code

Using Vector

Time Complexity: $O(n\log n)$, Space Complexity: $O(n)$

class Solution {
public:
    vector<int> findRightInterval(vector<vector<int>>& intervals) {
 
        // map of the start_i to index i, since 
        // start_i is unique
        unordered_map<int, int> mp;
        for(int i = 0; i < intervals.size(); i++) {
            mp[intervals[i][0]] = i;
        }
 
        sort(intervals.begin(), intervals.end());
 
        int n = intervals.size();
        vector<int> res(n, 0);
        for(int i = 0; i < n; i++) {
            int l = i, r = n - 1;
            
            int start_i = intervals[i][0];
            int end_i = intervals[i][1];
            // find end_i <= start_j
            while(l < r) {
                int m = l + (r - l) / 2;
                int start_j = intervals[m][0];
                if(start_j < end_i) {
                    l = m + 1;
                } else {
                    // start_j >= end_j
                    r = m;
                }
            } 
 
            res[mp[start_i]] = intervals[l][0] >= end_i ? mp[intervals[l][0]] : -1;
        }
        return res;
    }
};

Using Map

Time Complexity: $O(n\log n)$, Space Complexity: $O(n)$

用 Map 就等於使用 Vector 加上 unordered_map 去紀錄 index。共通點是都有使用到 lower_bound。

class Solution {
public:
    vector<int> findRightInterval(vector<vector<int>>& intervals) {
        int n = intervals.size();
        vector<int> res(n);
 
        map<int, int> mp; // <start_i, index>
        for(int i = 0; i < intervals.size(); i++) {
            mp[intervals[i][0]] = i;
        }
 
        for(int i = 0; i < intervals.size(); i++) {
		    // find start_i >= end_j(intervals[i][1])
            auto iter = mp.lower_bound(intervals[i][1]);
            res[i] = iter == mp.end() ? -1 : iter->second;
        }
 
        return res;
    }
};

Source

• Find Right Interval - LeetCode