LeetCode #2860 — MEDIUM

Happy Students

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[1,1]

Example 2

[6,0,3,3,6,7,2,7]

Step 02

Core Insight

What unlocks the optimal approach

If a student with <code>nums[i] = x</code> is selected, all the students with <code>nums[j] <= x</code> must be selected.
If a student with <code>nums[i] = x</code> is not selected, all the students with <code>nums[j] >= x</code> must not be selected.
Sort values in <code>nums</code> and try all possible values for <code>x</code> from <code>0</code> to <code>n</code> separately.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2860: Happy Students
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;
    }
}

// Accepted solution for LeetCode #2860: Happy Students
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
}

# Accepted solution for LeetCode #2860: Happy Students
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

// Accepted solution for LeetCode #2860: Happy Students
/**
 * [2860] Happy Students
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn count_ways(nums: Vec<i32>) -> i32 {
        let mut nums: Vec<usize> = nums.iter().map(|x| *x as usize).collect();
        let n = nums.len();
        nums.sort();

        // 是否能不选中任何学生
        let mut result = 0;

        for i in 0..=n {
            if i > 0 && nums[i - 1] >= i {
                continue;
            }

            if i < n && nums[i] <= i {
                continue;
            }

            result += 1;
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2860() {
        assert_eq!(2, Solution::count_ways(vec![1, 1]));
        assert_eq!(3, Solution::count_ways(vec![6, 0, 3, 3, 6, 7, 2, 7]));
        assert_eq!(1, Solution::count_ways(vec![1, 0, 1, 1]));
    }
}

// Accepted solution for LeetCode #2860: Happy Students
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.