Scale Engineer

Happy Students

Med
#2557Time: 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: O(n) to store the current subset being processed.
Patterns
Enumeration
Algorithms
Sorting
Data structures
Array

Prompt

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

Approaches

2 approaches with complexity analysis and trade-offs.

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.

Walkthrough

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:

  1. Calculate the size of the group, k = |S|.
  2. Initialize a flag all_happy = true.
  3. 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.
  4. 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

Time

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

O(n) to store the current subset being processed.

Trade-offs

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.

Solutions

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 walkthrough

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