Method 1 - Using Iterative Formula

First, sort all the grades. Then, assign 1 student to the first group, 2 students to the second group, and so on. This ensures that the i-th group is smaller in size and sum compared to the (i+1)-th group.

To determine the maximum possible k such that 1 + 2 + ... + k <= n:

Proof: When we sort the groups by size, the first group must have at least 1 student, the second group at least 2 students, and the k-th group at least k students.

This demonstrates the upper limit of k, and I have provided a method to achieve this arrangement.

Code

[{"language":"Java","code":"public int maximumGroups(int[] grades) {\n\tint k = 0, total = 0, n = grades.length;\n\twhile (total + k + 1<= n) {\n\t\tk++;\n\t\ttotal += k;\n\t}\n\treturn k;\n}","lang":"java","html":"<pre class=\"shiki shiki-themes github-light github-dark\" style=\"background-color:#fff;--shiki-dark-bg:#24292e;color:#24292e;--shiki-dark:#e1e4e8\" tabindex=\"0\"><code><span class=\"line\"><span style=\"color:#D73A49;--shiki-dark:#F97583\">public</span><span style=\"color:#D73A49;--shiki-dark:#F97583\"> int</span><span style=\"color:#6F42C1;--shiki-dark:#B392F0\"> maximumGroups</span><span style=\"color:#24292E;--shiki-dark:#E1E4E8\">(</span><span style=\"color:#D73A49;--shiki-dark:#F97583\">int</span><span style=\"color:#24292E;--shiki-dark:#E1E4E8\">[] grades) {</span></span>\n<span class=\"line\"><span style=\"color:#D73A49;--shiki-dark:#F97583\">\tint</span><span style=\"color:#24292E;--shiki-dark:#E1E4E8\"> k </span><span style=\"color:#D73A49;--shiki-dark:#F97583\">=</span><span style=\"color:#005CC5;--shiki-dark:#79B8FF\"> 0</span><span style=\"color:#24292E;--shiki-dark:#E1E4E8\">, total </span><span style=\"color:#D73A49;--shiki-dark:#F97583\">=</span><span style=\"color:#005CC5;--shiki-dark:#79B8FF\"> 0</span><span style=\"color:#24292E;--shiki-dark:#E1E4E8\">, n </span><span style=\"color:#D73A49;--shiki-dark:#F97583\">=</span><span style=\"color:#24292E;--shiki-dark:#E1E4E8\"> grades.length;&#x

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