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You are given a positive integer array `grades`

which represents the grades of students in a university. You would like to enter **all** these students into a competition in **ordered** non-empty groups, such that the ordering meets the following conditions:

- The sum of the grades of students in the
`i`

group is^{th}**less than**the sum of the grades of students in the`(i + 1)`

group, for all groups (except the last).^{th} - The total number of students in the
`i`

group is^{th}**less than**the total number of students in the`(i + 1)`

group, for all groups (except the last).^{th}

Return *the maximum number of groups that can be formed*.

**Example 1:**

```
Input: grades = [10,6,12,7,3,5]
Output: 3
Explanation: The following is a possible way to form 3 groups of students:
- 1st group has the students with grades = [12]. Sum of grades: 12. Student count: 1
- 2nd group has the students with grades = [6,7]. Sum of grades: 6 + 7 = 13. Student count: 2
- 3rd group has the students with grades = [10,3,5]. Sum of grades: 10 + 3 + 5 = 18. Student count: 3
It can be shown that it is not possible to form more than 3 groups.
```

**Example 2:**

```
Input: grades = [8,8]
Output: 1
Explanation: We can only form 1 group, since forming 2 groups would lead to an equal number of students in both groups.
```

**Constraints:**

`1 <= grades.length <= 10`

^{5}`1 <= grades[i] <= 10`

^{5}

```
public int maximumGroups(int[] grades) {
int k = 0, total = 0, n = grades.length;
while (total + k + 1 <= n)
total += ++k;
return k;
}
```

```
int maximumGroups(vector<int>& grades) {
int k = 0, total = 0, n = grades.size();
while (total + k + 1 <= n)
total += ++k;
return k;
}
```

```
def maximumGroups(self, A: List[int]) -> int:
n = len(A)
k = 0
while n >= k + 1:
k += 1
n -= k
return k
```

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