Question

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

Solution 1

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

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;
  }
}

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;
  }
};

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
}