README_EN.md

757. Set Intersection Size At Least Two

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Description

You are given a 2D integer array intervals where intervals[i] = [starti, endi] represents all the integers from starti to endi inclusively.

A containing set is an array nums where each interval from intervals has at least two integers in nums.

• For example, if intervals = [[1,3], [3,7], [8,9]], then [1,2,4,7,8,9] and [2,3,4,8,9] are containing sets.

Return the minimum possible size of a containing set.

 

Example 1:

Input: intervals = [[1,3],[3,7],[8,9]]
Output: 5
Explanation: let nums = [2, 3, 4, 8, 9].
It can be shown that there cannot be any containing array of size 4.

Example 2:

Input: intervals = [[1,3],[1,4],[2,5],[3,5]]
Output: 3
Explanation: let nums = [2, 3, 4].
It can be shown that there cannot be any containing array of size 2.

Example 3:

Input: intervals = [[1,2],[2,3],[2,4],[4,5]]
Output: 5
Explanation: let nums = [1, 2, 3, 4, 5].
It can be shown that there cannot be any containing array of size 4.

 

Constraints:

• 1 <= intervals.length <= 3000
• intervals[i].length == 2
• 0 <= starti < endi <= 108

Solutions

Python3

class Solution:
    def intersectionSizeTwo(self, intervals: List[List[int]]) -> int:
        intervals.sort(key=lambda x: (x[1], -x[0]))
        s = e = -1
        ans = 0
        for a, b in intervals:
            if a <= s:
                continue
            if a > e:
                ans += 2
                s, e = b - 1, b
            else:
                ans += 1
                s, e = e, b
        return ans

Java

class Solution {
    public int intersectionSizeTwo(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[1] == b[1] ? b[0] - a[0] : a[1] - b[1]);
        int ans = 0;
        int s = -1, e = -1;
        for (int[] v : intervals) {
            int a = v[0], b = v[1];
            if (a <= s) {
                continue;
            }
            if (a > e) {
                ans += 2;
                s = b - 1;
                e = b;
            } else {
                ans += 1;
                s = e;
                e = b;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int intersectionSizeTwo(vector<vector<int>>& intervals) {
        sort(intervals.begin(), intervals.end(), [&](vector<int>& a, vector<int>& b) {
            return a[1] == b[1] ? a[0] > b[0] : a[1] < b[1];
        });
        int ans = 0;
        int s = -1, e = -1;
        for (auto& v : intervals) {
            int a = v[0], b = v[1];
            if (a <= s) continue;
            if (a > e) {
                ans += 2;
                s = b - 1;
                e = b;
            } else {
                ans += 1;
                s = e;
                e = b;
            }
        }
        return ans;
    }
};

Go

func intersectionSizeTwo(intervals [][]int) int {
    sort.Slice(intervals, func(i, j int) bool {
        a, b := intervals[i], intervals[j]
        if a[1] == b[1] {
            return a[0] > b[0]
        }
        return a[1] < b[1]
    })
    ans := 0
    s, e := -1, -1
    for _, v := range intervals {
        a, b := v[0], v[1]
        if a <= s {
            continue
        }
        if a > e {
            ans += 2
            s, e = b - 1, b
        } else {
            ans += 1
            s, e = e, b
        }
    }
    return ans
}

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