Problem Description

You have an array grades representing positive integer grades of students in a university. Your task is to divide all these students into ordered groups for a competition, where each group must be non-empty.

The groups must satisfy two strict ordering conditions:

1. Grade Sum Condition: The sum of grades in group i must be less than the sum of grades in group (i+1) for all groups except the last one. This means each subsequent group must have a strictly higher total score.

2. Group Size Condition: The number of students in group i must be less than the number of students in group (i+1) for all groups except the last one. This means each subsequent group must have strictly more students.

Your goal is to find the maximum number of groups that can be formed while satisfying both conditions.

For example, if you have students with grades [1, 2, 3, 4, 5, 6], you could form groups like:

• Group 1: 1 student (smallest grade)
• Group 2: 2 students (next smallest grades)
• Group 3: 3 students (remaining grades)

This would give you 3 groups total, where each group has more students and a higher total score than the previous one.

The key insight is that to maximize the number of groups, you want to use the minimum possible number of students in each group while still satisfying the increasing size constraint. This leads to groups of sizes 1, 2, 3, ..., k, where the sum 1 + 2 + 3 + ... + k must not exceed the total number of students n.

Intuition

To maximize the number of groups, we need to think about the minimum resources required for each group while satisfying both constraints.

Since each group must have strictly more students than the previous one, the smallest possible configuration would be groups of sizes: 1, 2, 3, 4, ..., k. This uses the minimum number of students for k groups.

For the grade sum constraint, we can be clever: if we sort students by grades in ascending order and assign them sequentially to groups, we naturally satisfy the sum requirement. Why? Because:

• Group 1 gets the 1 lowest-grade student
• Group 2 gets the next 2 lowest-grade students
• Group 3 gets the next 3 lowest-grade students
• And so on...

Since we're always taking larger groups from higher-grade students, the sum automatically increases with each group.

The key insight is that the actual grade values don't matter - we can always arrange students to satisfy the sum constraint by sorting. What limits us is the group size constraint.

With groups of sizes 1, 2, 3, ..., k, we need a total of 1 + 2 + 3 + ... + k = k*(k+1)/2 students. This must not exceed our total number of students n.

Therefore, we need to find the largest k such that k*(k+1)/2 ≤ n, which can be rewritten as k² + k ≤ 2n.

This is where binary search comes in - we're looking for the maximum value of k that satisfies this quadratic inequality. The solution uses bisect_right to find the largest k where k² + k ≤ 2n, which efficiently solves this mathematical optimization problem.

Learn more about Greedy, Math and Binary Search patterns.

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 use binary search to find this maximum k:

• Set left boundary l = 1 and right boundary r = n
• For each midpoint mid = ⌊(l + r + 1) / 2⌋:
  • If (1 + mid) × mid > 2 × n, then mid is too large, set r = mid - 1
  • Otherwise, mid is valid, set l = mid
• Return l as the answer

Implementation Details

The provided solution cleverly uses Python's bisect_right function:

bisect_right(range(n + 1), n * 2, key=lambda x: x * x + x) - 1

This finds the rightmost position where x² + x ≤ 2n in the range [0, n], then subtracts 1 to get the actual value of k.

The key=lambda x: x * x + x transforms each value x to x² + x for comparison with 2n.

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 We need to find the largest k where k² + k ≤ 2n = 16

Let's trace through the binary search:

• Initial range: l = 1, r = 8
• First iteration: mid = (1 + 8 + 1) / 2 = 5
  • Check: 5² + 5 = 30 > 16, so 5 is too large
  • Update: r = 4
• Second iteration: mid = (1 + 4 + 1) / 2 = 3
  • Check: 3² + 3 = 12 ≤ 16, so 3 is valid
  • Update: l = 3
• Third iteration: mid = (3 + 4 + 1) / 2 = 4
  • Check: 4² + 4 = 20 > 16, so 4 is too large
  • Update: r = 3
• Loop ends since l = r = 3

Result: k = 3 (we can form 3 groups)

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.

Solution Implementation

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: 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.

# Incorrect approach - trying to actually form groups
def maximumGroups_wrong(grades):
    grades.sort()
    groups = []
    start = 0
    group_size = 1
  
    while start < len(grades):
        if start + group_size > len(grades):
            break
        current_group = grades[start:start + group_size]
        if groups and sum(current_group) <= sum(groups[-1]):
            break  # This check is unnecessary!
        groups.append(current_group)
        start += group_size
        group_size += 1
  
    return len(groups)

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 2: Integer Overflow in Arithmetic

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

# Potential overflow issue
def maximumGroups_overflow(grades):
    n = len(grades)
    k = 1
    while k * (k + 1) // 2 <= n:  # k * (k + 1) might overflow
        k += 1
    return k - 1

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
while k * (k + 1) <= 2 * n:  # Less likely to overflow
    k += 1

Pitfall 3: Off-by-One Errors in Binary Search

The Mistake: Incorrectly implementing the binary search boundaries or forgetting to subtract 1 from the bisect_right result.

# Common mistakes with binary search
def maximumGroups_off_by_one(grades):
    n = len(grades)
  
    # Mistake 1: Wrong boundary initialization
    left, right = 0, n  # Should start left at 1, not 0
  
    # Mistake 2: Incorrect mid calculation
    while left < right:  # Should be left <= right
        mid = (left + right) // 2  # Should be (left + right + 1) // 2 for upper bound
        if mid * (mid + 1) <= 2 * n:
            left = mid  # This creates infinite loop without proper adjustment
        else:
            right = mid - 1
  
    return left

The Fix: Use the proven bisect_right approach or carefully implement binary search:

def maximumGroups_correct(grades):
    n = len(grades)
    left, right = 1, n
  
    while left < right:
        mid = (left + right + 1) // 2  # Round up for finding maximum
        if mid * (mid + 1) <= 2 * n:
            left = mid
        else:
            right = mid - 1
  
    return left

Pitfall 4: Misunderstanding the Problem Constraints

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

# Incorrect - analyzing grade values unnecessarily
def maximumGroups_overcomplex(grades):
    grades.sort()
    min_sum_difference = float('inf')
    for i in range(1, len(grades)):
        min_sum_difference = min(min_sum_difference, grades[i] - grades[i-1])
    # Trying to use min_sum_difference to calculate 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.