让所有学生保持开心的分组方法数

You are given a 0-indexed integer array nums of length n where n is the total number of students in the class. The class teacher tries to select a group of students so that all the students remain happy.

The ith student will become happy if one of these two conditions is met:

• The student is selected and the total number of selected students is strictly greater than nums[i].
• The student is not selected and the total number of selected students is strictly less than nums[i].

Return the number of ways to select a group of students so that everyone remains happy.

 

Example 1:

Input: nums = [1,1]
Output: 2
Explanation: 
The two possible ways are:
The class teacher selects no student.
The class teacher selects both students to form the group. 
If the class teacher selects just one student to form a group then the both students will not be happy. Therefore, there are only two possible ways.

Example 2:

Input: nums = [6,0,3,3,6,7,2,7]
Output: 3
Explanation: 
The three possible ways are:
The class teacher selects the student with index = 1 to form the group.
The class teacher selects the students with index = 1, 2, 3, 6 to form the group.
The class teacher selects all the students to form the group.

 

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] < nums.length

/*
 * 思路：
 * 1. 首先将数组 nums 进行排序。
 * 2. 使用 Java Stream 对数组进行处理。
 * 3. 计算有多少学生会保持开心，分为两种情况：
 *    a. 学生被选中且选中的学生人数大于 nums[i]。
 *    b. 学生未被选中且选中的学生人数小于 nums[i]。
 * 4. 遍历排序后的数组，统计满足条件的学生人数，并根据这些人数计算出满足所有学生开心的分组方法数。
 */
import java.util.Arrays;

public class Solution {
    public long countHappyGroups(int[] nums) {
        int n = nums.length;
        Arrays.sort(nums);
        return Arrays.stream(nums)
                     .boxed()
                     .collect(Collectors.groupingBy(i -> i))
                     .entrySet()
                     .stream()
                     .mapToInt(entry -> {
                         int count = entry.getValue().size();
                         return (count == 0 || nums[count - 1] < count) && (count == n || nums[count] > count) ? 1 : 0;
                     })
                     .sum();
    }
}

题解首先对数组 nums 进行排序。排序后，可以利用数组的顺序属性来判断哪些选中的学生人数可以使全部学生保持开心。\n\n对于每个学生 i ，要满足的条件是：\n1. 如果这位学生被选中，那么至少要有 nums[i] + 1 个学生被选中（因为要严格大于 nums[i]）。\n2. 如果这位学生没有被选中，被选中的学生数必须严格小于 nums[i]。\n\n由于数组已经排序，nums[i] 是递增的，因此可以通过遍历排序后的数组，检查每个可能的学生人数 k 是否满足所有学生的开心条件：\n- nums[0] > 0 意味着没有学生被选中时，所有学生都会开心。\n- 对于 k = i (从 1 开始)，如果 nums[i-1] < i < nums[i]，则选择 i 个学生会使得所有学生保持开心。\n\n最终，算法统计并返回所有可能让学生开心的组合数量。

O(n log n)

O(1)