LeetCode 2860. Happy Students Solution in Java, C++, Python & More | Explanation + Code

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 × log n), and the space complexity is O(log n). Here, n is the length of the array.

PythonJavaC++GoTypeScript

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 ans += 1 return ans(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)