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986. Interval List Intersections (Medium)

You are given two lists of closed intervals, firstList and secondList, where firstList[i] = [starti, endi] and secondList[j] = [startj, endj]. Each list of intervals is pairwise disjoint and in sorted order.

Return the intersection of these two interval lists.

A closed interval [a, b] (with a <= b) denotes the set of real numbers x with a <= x <= b.

The intersection of two closed intervals is a set of real numbers that are either empty or represented as a closed interval. For example, the intersection of [1, 3] and [2, 4] is [2, 3].

 

Example 1:

Input: firstList = [[0,2],[5,10],[13,23],[24,25]], secondList = [[1,5],[8,12],[15,24],[25,26]]
Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]

Example 2:

Input: firstList = [[1,3],[5,9]], secondList = []
Output: []

Example 3:

Input: firstList = [], secondList = [[4,8],[10,12]]
Output: []

Example 4:

Input: firstList = [[1,7]], secondList = [[3,10]]
Output: [[3,7]]

 

Constraints:

• 0 <= firstList.length, secondList.length <= 1000
• firstList.length + secondList.length >= 1
• 0 <= starti < endi <= 109
• endi < starti+1
• 0 <= startj < endj <= 109
• endj < startj+1

Companies:
Facebook, Uber, Yandex, Google, Amazon

Related Topics:
Array, Two Pointers

Similar Questions:

• Merge Intervals (Medium)
• Merge Sorted Array (Easy)
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Solution 1.

// OJ: https://leetcode.com/problems/interval-list-intersections/
// Author: github.com/lzl124631x
// Time: O(M+N)
// Space: O(1)
class Solution {
public:
    vector<vector<int>> intervalIntersection(vector<vector<int>>& A, vector<vector<int>>& B) {
        int M = A.size(), N = B.size();
        vector<vector<int>> ans;
        for (int i = 0, j = 0; i < M && j < N; ) {
            int s = max(A[i][0], B[j][0]), e = min(A[i][1], B[j][1]);
            if (s <= e) ans.push_back({ s, e });
            if (A[i][1] < B[j][1]) ++i;
            else ++j;
        }
        return ans;
    }
};