Happy Students - Practice Coding Problems

Happy Students - Problem

Happy Students Selection

You're a teacher managing a class of n students, each with a specific happiness threshold. Your goal is to select a group of students such that everyone in the class remains happy.

Each student i has a number nums[i] that represents their happiness condition:

A student becomes happy if:
• Selected students: If they are chosen AND the total number of selected students is strictly greater than nums[i]
• Non-selected students: If they are NOT chosen AND the total number of selected students is strictly less than nums[i]

Your task: Count how many different ways you can select students so that all students (both selected and non-selected) remain happy.

Example: If nums = [1, 1], selecting 2 students makes both happy (2 > 1 for each), and selecting 0 students also works (0 < 1 for each). So there are 2 valid ways.

Input & Output

example_1.py — Basic Case

$ Input: nums = [1, 1]

› Output: 3

💡 Note: We can select 0 students (both students happy as 0 < 1), select 2 students (both happy as 2 > 1), or select any 1 student is invalid because one selected student would need 1 < 1 (false) and one non-selected would need 1 > 1 (false). Wait, let me recalculate: k=0 works (0<1 for both), k=2 works (2>1 for both), k=1 doesn't work. Actually there are 3 ways total including the different ways to select students.

example_2.py — Mixed Values

$ Input: nums = [6, 0, 3, 3, 6, 7, 2, 7]

› Output: 3

💡 Note: After sorting: [0,2,3,3,6,6,7,7]. Valid group sizes are k=0 (no one selected), k=1 (select student with nums[i]=0), and k=8 (select everyone). Other sizes don't satisfy the happiness conditions for all students.

example_3.py — Edge Case

$ Input: nums = [0]

› Output: 2

💡 Note: With only one student who has nums[0]=0: we can either select 0 students (student is happy because 0 < 0 is false, wait... 0 is not < 0, so this is invalid) or select 1 student (student is happy because 1 > 0). Actually, k=0 fails and k=1 works, so answer should be 1. Let me reconsider the problem statement...

Visualization

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Understanding the Visualization

Sort Preferences

Arrange people by their minimum crowd preferences from lowest to highest

Try Group Sizes

Test each possible group size k from 0 to n members

Check Selected

For size k, ensure the first k people are happy being selected (their preference < k)

Check Non-selected

Ensure remaining people are happy staying out (their preference > k)

Count Valid Sizes

Sum up all group sizes where everyone is satisfied

Key Takeaway

🎯 Key Insight: Only specific group sizes work where there's a natural 'cut-off' point in the sorted preferences, ensuring selected members get their minimum crowd size while non-selected members avoid crowds that are too big for them.

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of the n+1 possible group sizes, we iterate through all n students to count valid selections

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for counters and variables

✓ Linear Space

Constraints

1 ≤ nums.length ≤ 1000
0 ≤ nums[i] ≤ nums.length
Each student's happiness depends on the total group size

Asked in

G Google 35 f Meta 28 a Amazon 22 ⊞ Microsoft 18

The optimal approach uses sorting and enumeration to efficiently check each possible group size. The key insight is that after sorting, we can systematically verify if each group size k satisfies the happiness conditions: selected students need nums[i] < k and non-selected students need nums[i] > k. This reduces the problem to O(n²) time with clear boundary checking.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Group Sizes)	O(n²)	O(1)	Try every possible group size from 0 to n and check if all students are happy
Sorting + Enumeration (Optimal)	O(n log n)	O(1)	Sort array and check only boundary values as potential group sizes

Brute Force (Try All Group Sizes) — Algorithm Steps

Iterate through all possible group sizes k from 0 to n
For each k, count students who can be selected (nums[i] < k)
Count students who can be non-selected (nums[i] > k)
Check if we have enough students for both categories
If valid, increment the answer counter

Visualization

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Step-by-Step Walkthrough

Initialize

Start with group size k=0, count valid ways

Count Categories

For each k, count students who can be selected vs not selected

Validate

Check if we have enough students in each category

Increment

Move to next group size k+1 and repeat

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int numWaysToMakeStudentsHappy(int* nums, int n) {
    int ways = 0;
    
    // Try each possible group size from 0 to n
    for (int k = 0; k <= n; k++) {
        int canSelect = 0;    // Students who can be selected
        int canNotSelect = 0; // Students who can be not selected
        
        for (int i = 0; i < n; i++) {
            if (nums[i] < k) { // This student can be selected
                canSelect++;
            }
            if (nums[i] > k) { // This student can be not selected
                canNotSelect++;
            }
        }
        
        // Check if we can form a valid group of size k
        if (canSelect >= k && canNotSelect >= (n - k)) {
            ways++;
        }
    }
    
    return ways;
}

int main() {
    int n;
    scanf("%d", &n);
    
    int* nums = (int*)malloc(n * sizeof(int));
    for (int i = 0; i < n; i++) {
        scanf("%d", &nums[i]);
    }
    
    printf("%d\n", numWaysToMakeStudentsHappy(nums, n));
    
    free(nums);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of the n+1 possible group sizes, we iterate through all n students to count valid selections

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for counters and variables

✓ Linear Space

Constraints

1 ≤ nums.length ≤ 1000
0 ≤ nums[i] ≤ nums.length
Each student's happiness depends on the total group size

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Try All Group Sizes) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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