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2860. Happy Students
Description
You are given a 0indexed 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 i^{th}
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 <= 10^{5}
0 <= nums[i] < nums.length
Solutions
Solution 1: Sorting + Enumeration
Assume that $k$ students are selected, then the following conditions hold:
 If $nums[i] = k$, then there is no grouping method;
 If $nums[i] > k$, then student $i$ is not selected;
 If $nums[i] < k$, then student $i$ is selected.
Therefore, the selected students must be the first $k$ elements in the sorted $nums$ array.
We enumerate $k$ in the range $[0,..n]$. For the current number of selected students $i$, we can get the maximum student number in the group $i1$, which is $nums[i1]$. If $i > 0$ and $nums[i1] \ge i$, then there is no grouping method; if $i < n$ and $nums[i] \le i$, then there is no grouping method. Otherwise, there is a grouping method, and the answer is increased by one.
After the enumeration ends, return the answer.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array.

class Solution { public int countWays(List<Integer> nums) { Collections.sort(nums); int n = nums.size(); int ans = 0; for (int i = 0; i <= n; i++) { if ((i == 0  nums.get(i  1) < i) && (i == n  nums.get(i) > i)) { ans++; } } return ans; } }

class Solution { public: int countWays(vector<int>& nums) { sort(nums.begin(), nums.end()); int ans = 0; int n = nums.size(); for (int i = 0; i <= n; ++i) { if ((i && nums[i  1] >= i)  (i < n && nums[i] <= i)) { continue; } ++ans; } return ans; } };

class Solution: def countWays(self, nums: List[int]) > int: nums.sort() n = len(nums) ans = 0 for i in range(n + 1): if i and nums[i  1] >= i: continue if i < n and nums[i] <= i: continue return ans

func countWays(nums []int) (ans int) { sort.Ints(nums) n := len(nums) for i := 0; i <= n; i++ { if (i > 0 && nums[i1] >= i)  (i < n && nums[i] <= i) { continue } ans++ } return }

function countWays(nums: number[]): number { nums.sort((a, b) => a  b); let ans = 0; const n = nums.length; for (let i = 0; i <= n; ++i) { if ((i && nums[i  1] >= i)  (i < n && nums[i] <= i)) { continue; } ++ans; } return ans; }