986. Interval List Intersections

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https://leetcode.com/problems/interval-list-intersections/

Given two lists of closed intervals, each list of intervals is pairwise disjoint and in sorted order.
Return the intersection of these two interval lists.
(Formally, 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 is either empty, or can be represented as a closed interval.  For example, the intersection of [1, 3] and [2, 4] is [2, 3].)

Example 1:
Input: A = [[0,2],[5,10],[13,23],[24,25]], B = [[1,5],[8,12],[15,24],[25,26]]
Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]
Reminder: The inputs and the desired output are lists of Interval objects, and not arrays or lists.

Note:
  1. 0 <= A.length < 1000
  2. 0 <= B.length < 1000
  3. 0 <= A[i].start, A[i].end, B[i].start, B[i].end < 10^9
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Clarifying questions
Are input lists sorted
What is expected output when start, end times overlap
What is expected output when any of the input list is empty
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Intuition

Find overlaps with simple formula

lo = Math.max(A.start, B.start)
hi = Math.min(A.end, B.end)

if (lo <= hi) => valid overlap -- add to ans

Increment the pointer of array with min end time, it no longer contributes to ans
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Time - O(A + B)
Space - O(1) - not counting space for ans
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Related problems
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class Solution {
public int[][] intervalIntersection(int[][] A, int[][] B) {
List<int[]> ans = new ArrayList<int[]>();
int i = 0, j = 0;
while (i < A.length && j < B.length) {
int lo = Math.max(A[i][0], B[j][0]);
int hi = Math.min(A[i][1], B[j][1]);
// Check valid overlap
if (lo <= hi) {
ans.add(new int[] {lo, hi});
}
// Advance the ended interval, it cannot add to output
if (A[i][1] < B[j][1]) {
i++;
} else {
j++;
}
}
return ans.toArray(new int[ans.size()][2]);
}
}
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