Interval List Intersections

May 23, 2020

Introduction

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 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:

0 <= A.length < 1000
0 <= B.length < 1000
0 <= A[i].start, A[i].end, B[i].start, B[i].end < 10^9
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

Solution

Let’s do merge of two sorted arrays, keeping in mind about intervals and use min and max methods to test them.

func intervalIntersection(A [][]int, B [][]int) [][]int {
    var out [][]int
    for i, j := 0, 0; i < len(A) && j < len(B); {
        x := max(A[i][0], B[j][0])
        y := min(A[i][1], B[j][1])
        if x <= y {
            out = append(out, []int { x, y })
        }
        if A[i][1] <= y {
            i++
        } else if A[i][0] <= y {
            A[i][0] = y+1  
        }
        if B[j][1] <= y {
            j++
        } else if B[j][0] <= y {
            B[j][0] = y+1
        }
    }
    return out
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

Performance of this solution

Runtime: 20 ms
Memory Usage: 6.3 MB