Question

2860. Happy Students

中文文档

Description

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

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 $i-1$, which is $nums[i-1]$. If $i > 0$ and $nums[i-1] \ge i$, 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:
  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

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;
  }
};

func countWays(nums []int) (ans int) {
  sort.Ints(nums)
  n := len(nums)
  for i := 0; i <= n; i++ {
    if (i > 0 && nums[i-1] >= 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;
}