Solution Approach

The solution uses a greedy approach combined with binary search to find the maximum number of groups.

Step 1: Sort and Assign Strategy

First, we sort the students by their grades in ascending order. Then we assign students to groups sequentially:

• Group 1: 1 student (lowest grade)
• Group 2: 2 students (next 2 lowest grades)
• Group 3: 3 students (next 3 lowest grades)
• Continue this pattern...

This greedy assignment ensures both conditions are met:

• Each group has more students than the previous one
• Each group has a higher sum than the previous one (since we're taking more students from higher grades)

Step 2: Finding Maximum k Using Binary Search

The problem reduces to finding the largest k such that: 1 + 2 + 3 + ... + k ≤ n

Using the formula for arithmetic series: (1 + k) × k / 2 ≤ n

Rearranging: k² + k ≤ 2n

We define our feasible function: feasible(k) = true if k² + k <= 2n. This creates a pattern like [true, true, true, false, false, ...] as k increases. We want to find the last true value.

However, to use the standard binary search template (which finds the first true), we can invert the condition: not_feasible(k) = k² + k > 2n, which creates [false, false, false, true, true, ...]. The first true gives us k+1 where k is our answer.

Using the standard template:

left, right = 1, n
first_true_index = n + 1  # Default if all are feasible

while left <= right:
    mid = (left + right) // 2
    if mid * mid + mid > 2 * n:  # Not feasible (too many groups)
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1

return first_true_index - 1  # The last feasible k

Time Complexity: O(log n) for binary search Space Complexity: O(1) as we only use a few variables

Example Walkthrough

Let's walk through the solution with a concrete example: grades = [8, 2, 3, 1, 5, 6, 9, 7] (n = 8 students).

Step 1: Sort the grades After sorting: [1, 2, 3, 5, 6, 7, 8, 9]

Step 2: Find maximum k using binary search template We need to find the largest k where k² + k ≤ 2n = 16. Using the inverted condition (finding first k where k² + k > 16):

Initialize: left = 1, right = 8, first_true_index = 9 (default)

Iteration 1: left = 1, right = 8

• mid = (1 + 8) // 2 = 4
• Check: 4² + 4 = 20 > 16 → not feasible (true for inverted condition)
• Update first_true_index = 4, right = mid - 1 = 3

Iteration 2: left = 1, right = 3

• mid = (1 + 3) // 2 = 2
• Check: 2² + 2 = 6 ≤ 16 → feasible (false for inverted condition)
• Update left = mid + 1 = 3

Iteration 3: left = 3, right = 3

• mid = (3 + 3) // 2 = 3
• Check: 3² + 3 = 12 ≤ 16 → feasible (false for inverted condition)
• Update left = mid + 1 = 4

Iteration 4: left = 4, right = 3

• left > right, loop terminates
• first_true_index = 4 (first k where condition fails)
• Result: k = first_true_index - 1 = 3

Step 3: Form the groups With k = 3, we form groups of sizes 1, 2, and 3:

• Group 1: [1] (1 student, sum = 1)
• Group 2: [2, 3] (2 students, sum = 5)
• Group 3: [5, 6, 7] (3 students, sum = 18)

Note: We use 1 + 2 + 3 = 6 students total, leaving 2 students unused [8, 9].

Verification:

• Size constraint: 1 < 2 < 3 ✓
• Sum constraint: 1 < 5 < 18 ✓

Both conditions are satisfied, and we've formed the maximum possible 3 groups.

Time and Space Complexity

The time complexity is O(log n) where n is the total number of students.

The bisect_right function performs a binary search on the range range(n + 1), which creates a virtual sequence of numbers from 0 to n. The binary search algorithm divides the search space in half with each iteration, resulting in logarithmic time complexity. The key function lambda x: x * x + x is evaluated O(log n) times during the binary search, and each evaluation takes O(1) time.

The space complexity is O(1).

Although range(n + 1) is used as an argument, Python's range object is a lazy iterator that doesn't create the entire list in memory - it generates values on demand. The bisect_right function only needs to maintain a few variables (left pointer, right pointer, mid pointer) during the binary search process, which requires constant space. The lambda function also uses constant space for its computation.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Using Wrong Binary Search Template

The Mistake: Using while left < right with left = mid pattern instead of the standard template with while left <= right, first_true_index, and right = mid - 1.

Why It's Wrong: Different binary search variants behave differently at boundaries. Using inconsistent patterns makes code harder to reason about and more prone to off-by-one errors.

The Fix: Use the standard template with inverted condition:

left, right = 1, n
first_true_index = n + 1  # Default value

while left <= right:
    mid = (left + right) // 2
    if mid * mid + mid > 2 * n:  # Inverted: first k that doesn't work
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1

return first_true_index - 1

Pitfall 2: Attempting to Actually Form the Groups

The Mistake: Many developers initially try to sort the array and explicitly form the groups by partitioning the students, tracking sums, and validating conditions. This leads to unnecessary complexity and potential errors.

Why It's Wrong: The problem guarantees that if we sort the grades and assign students to groups of sizes 1, 2, 3, ..., k, the sum condition will automatically be satisfied. We don't need to validate it explicitly.

The Fix: Recognize that this is purely a mathematical problem about partitioning n items into groups of sizes 1, 2, 3, ..., k.

Pitfall 3: Integer Overflow in Arithmetic

The Mistake: When calculating k * (k + 1), intermediate results can overflow for large values of n.

Why It's Wrong: For very large n (close to integer limits), k * (k + 1) might exceed the maximum integer value before division.

The Fix: Rearrange the comparison to avoid overflow:

# Better approach - comparing with 2*n instead
if mid * mid + mid > 2 * n:  # Less likely to overflow

Pitfall 4: Misunderstanding the Problem Constraints

The Mistake: Assuming that the actual grade values matter for determining the maximum number of groups.

Why It's Wrong: The maximum number of groups depends only on the total number of students (n), not on the specific grade values. The grades just need to be positive integers.

The Fix: Focus only on the count of students and the mathematical relationship between group sizes and total students.