LeetCode 2860 - Happy Students

The problem asks us to determine the number of ways to select a subset of students from a class such that every student is happy according to the given rules. Each student has a number nums[i] that represents a threshold.

leetcodemediumarraysortingenumeration

LeetCode Problem 2860

Difficulty: 🟡 Medium
Topics: Array, Sorting, Enumeration

Solution

Problem Understanding

The problem asks us to determine the number of ways to select a subset of students from a class such that every student is happy according to the given rules. Each student has a number nums[i] that represents a threshold. A student is happy if they are selected and the total number of selected students is strictly greater than nums[i], or if they are not selected and the total number of selected students is strictly less than nums[i].

The input is a 0-indexed integer array nums of length n, representing each student’s happiness threshold. The output is a single integer representing the number of valid selections of students.

Constraints are important for understanding the potential scale of the solution. With 1 <= nums.length <= 10^5 and 0 <= nums[i] < nums.length, a brute-force approach that checks every possible subset of students would be infeasible, since there are 2^n subsets. Instead, we need a more efficient method. Important edge cases include arrays where all values are the same, arrays where values are at their extremes (0 or n-1), and arrays of minimal or maximal length.

Approaches

The brute-force approach would involve iterating through every possible subset of students and checking if each selection satisfies the happiness condition for all students. While correct, this approach has exponential time complexity O(2^n) and is not feasible for large n up to 10^5.

The key insight for an optimal solution is to recognize that the problem can be simplified by sorting the array. Once nums is sorted, the number of students selected can be considered in increasing order from 0 to n. For a given k selected students, all students with nums[i] < k must be selected and all students with nums[i] > k must not be selected. This allows us to check each potential k efficiently and count valid configurations without enumerating all subsets.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n * n) O(n) Check all subsets explicitly; infeasible for large n
Optimal O(n log n) O(1) or O(n) Sort array and check valid selection counts sequentially

Algorithm Walkthrough

  1. Sort the input array nums in ascending order. Sorting ensures we can reason about thresholds in order, which is key for counting valid groups efficiently.
  2. Initialize a counter ways to 0. This will accumulate the number of valid selections.
  3. Iterate over all possible selection sizes k from 0 to n. For each k, check if it forms a valid selection:
  • If k == 0, it is valid if the smallest threshold nums[0] > 0.
  • If k == n, it is valid if the largest threshold nums[n-1] < n.
  • Otherwise, for 0 < k < n, check if nums[k-1] < k and nums[k] > k. This ensures students who need fewer selections than k are happy, and those who need more than k are not selected.
  1. Increment ways for each valid k.
  2. Return ways as the result.

Why it works: Sorting allows us to identify boundaries where selected and unselected students meet the happiness condition. The check nums[k-1] < k and nums[k] > k guarantees that all selected students have thresholds below k and all unselected students have thresholds above k, thus ensuring happiness.

Python Solution

from typing import List

class Solution:
    def countWays(self, nums: List[int]) -> int:
        nums.sort()
        n = len(nums)
        ways = 0
        
        for k in range(n + 1):
            if k == 0:
                if nums[0] > 0:
                    ways += 1
            elif k == n:
                if nums[-1] < n:
                    ways += 1
            else:
                if nums[k-1] < k and nums[k] > k:
                    ways += 1
                    
        return ways

The Python implementation follows the algorithm directly. We sort nums first, iterate through possible selection sizes k, and use conditions to validate if a selection size is feasible. Edge cases for selecting no students and all students are handled separately to avoid index errors.

Go Solution

import "sort"

func countWays(nums []int) int {
    sort.Ints(nums)
    n := len(nums)
    ways := 0

    for k := 0; k <= n; k++ {
        if k == 0 {
            if nums[0] > 0 {
                ways++
            }
        } else if k == n {
            if nums[n-1] < n {
                ways++
            }
        } else {
            if nums[k-1] < k && nums[k] > k {
                ways++
            }
        }
    }

    return ways
}

The Go version mirrors the Python solution. Sorting is done using sort.Ints. The iteration logic handles edge cases at k=0 and k=n separately. Go handles slices and indexing safely without needing additional checks, and integer overflow is not a concern due to constraints.

Worked Examples

Example 1: nums = [1,1]

After sorting: [1,1]

Possible k values and checks:

k Condition Valid
0 nums[0] > 0 → 1 > 0 Yes
1 nums[0] < 1 and nums[1] > 1 → 1 < 1 and 1 > 1 No
2 nums[-1] < 2 → 1 < 2 Yes

Result: 2 ways.

Example 2: nums = [6,0,3,3,6,7,2,7]

After sorting: [0,2,3,3,6,6,7,7]

Valid k values: 1, 4, 8

Result: 3 ways.

Complexity Analysis

Measure Complexity Explanation
Time O(n log n) Sorting dominates, iterating through k is O(n)
Space O(1) In-place sorting; additional variables are O(1)

Sorting is the only step that exceeds linear time, and we do not use extra space beyond a few integers for counting and indexing.

Test Cases

# Basic examples
assert Solution().countWays([1,1]) == 2  # two ways: no student or all students
assert Solution().countWays([6,0,3,3,6,7,2,7]) == 3  # three valid selection sizes

# Edge cases
assert Solution().countWays([0]) == 1  # single student with 0 threshold
assert Solution().countWays([0,0,0]) == 1  # only selecting all students works
assert Solution().countWays([2,2,2]) == 1  # only selecting no students works
assert Solution().countWays([0,1,2,3,4]) == 2  # either none or all selected
assert Solution().countWays([4,4,4,4,4]) == 1  # only none selected works
Test Why
[1,1] Basic scenario with small array
[6,0,3,3,6,7,2,7] Random values, tests general algorithm
[0] Minimal input, edge of single student
[0,0,0] All thresholds zero, selection constraints test
[2,2,2] All thresholds equal, selecting none is the only way
[0,1,2,3,4] Increasing sequence, tests selection boundaries
[4,4,4,4,4] All high thresholds, only none selected works

Edge Cases

One important edge case is when all students have the same threshold. This tests whether the implementation correctly identifies the only valid selection (all or none). Another edge case is when the thresholds include both extremes, 0 and n-1, which tests boundary conditions for k=0 and k=n. Finally, an edge case with a single student must be handled correctly, ensuring that the selection conditions are not misapplied. Each of these cases is handled in the implementation by separate checks for k=0 and k=n and by the sorted array comparisons for general k.