Happy Students

MEDIUM

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 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 <= 10⁵
0 <= nums[i] < nums.length

Approaches

Checkout 2 different approaches to solve Happy Students. Click on different approaches to view the approach and algorithm in detail.

Brute Force by Checking All Subsets

This approach exhaustively checks every possible group of students. A group can be represented as a subset of the students. There are 2^n possible subsets, where n is the number of students. For each subset, we determine its size k and then verify if every student (both in the subset and not in the subset) is happy according to the given conditions.

Algorithm

Initialize a counter ways to 0.
Generate every possible subset of students. There are 2^n such subsets.
For each subset S:
- Let k be the size of the subset S.
- Assume the group is valid (all_happy = true).
- Iterate through all n students:
  - If a student i is in S (selected), check if k > nums[i]. If not, the group is invalid (all_happy = false), and we can stop checking this subset.
  - If a student i is not in S (not selected), check if k < nums[i]. If not, the group is invalid (all_happy = false), and we can stop checking this subset.
- If all_happy remains true after checking all students, increment ways.
Return ways.

The algorithm iterates through all 2^n subsets of the n students. This can be done using recursion or iteration with bit manipulation (where the i-th bit of a number from 0 to 2^n - 1 represents whether student i is selected).

For each subset S:

Calculate the size of the group, k = |S|.
Initialize a flag all_happy = true.
Iterate through all n students from i = 0 to n-1:
- If student i is in the subset S (selected): check if k > nums[i]. If not, set all_happy = false and break.
- If student i is not in the subset S (not selected): check if k < nums[i]. If not, set all_happy = false and break.
If all_happy is still true after checking all students, it means this subset forms a valid group, so we increment our total count of ways.

After checking all 2^n subsets, the final count is the answer.

Complexity Analysis

Time Complexity: O(n * 2^n). There are `2^n` subsets to check. For each subset, we iterate through all `n` students to verify the happiness conditions.Space Complexity: O(n) to store the current subset being processed.

Pros and Cons

Pros:

Conceptually simple and directly follows the problem definition.

Cons:

Extremely inefficient due to its exponential time complexity.
Not feasible for the given constraints where n can be up to 10^5.

Code Solutions

Checking out 3 solutions in different languages for Happy Students. Click on different languages to view the code.

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;
  }
}

Video Solution

Watch the video walkthrough for Happy Students

Algorithms:

Sorting

Patterns:

Enumeration

Data Structures:

Array