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.

Approach #1:

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    vector<int> findRightInterval(vector<Interval>& intervals) {
        int len = intervals.size();
        vector<int> ans;
        map<int, int> temp;
        for (int i = 0; i < len; ++i) {
            temp[intervals[i].start] = i;
        }
        for (int i = 0; i < len; ++i) {
            auto it = temp.lower_bound(intervals[i].end);
            if (it != temp.end()) ans.push_back(it->second);
            else ans.push_back(-1);
        }
        return ans;
    }
};

Runtime: 64 ms, faster than 69.43% of C++ online submissions for Find Right Interval.

 Analysis:

std::map::lower_bound

      iterator lower_bound (const key_type& k);
const_iterator lower_bound (const key_type& k) const;

Return iterator to lower bound

Returns an iterator pointing to the first element in the container whose key is not considered to go before k (i.e., either it is equivalent or goes after).

The function uses its internal comparison object (key_comp) to determine this, returning an iterator to the first element for which key_comp(element_key,k) would return false.

If the map class is instantiated with the default comparison type (less), the function returns an iterator to the first element whose key is not less than k.

A similar member function, upper_bound, has the same behavior as lower_bound, except in the case that the mapcontains an element with a key equivalent to k: In this case, lower_bound returns an iterator pointing to that element, whereas upper_bound returns an iterator pointing to the next element.

Parameters

k

Key to search for.
Member type key_type is the type of the elements in the container, defined in map as an alias of its first template parameter (Key).

Return value

An iterator to the the first element in the container whose key is not considered to go before k, or map::end if all keys are considered to go before k.

If the map object is const-qualified, the function returns a const_iterator. Otherwise, it returns an iterator.

Member types iterator and const_iterator are bidirectional iterator types pointing to elements (of type value_type).
Notice that value_type in map containers is itself also a pair type: pair<const key_type, mapped_type>.