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Formatted question description: https://leetcode.ca/all/2358.html

# 2358. Maximum Number of Groups Entering a Competition

• Difficulty: Medium.
• Related Topics: Array, Math, Binary Search, Greedy.
• Similar Questions: Maximum Height by Stacking Cuboids .

## Problem

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 ith group is less than the sum of the grades of students in the (i + 1)th group, for all groups (except the last).

• The total number of students in the ith group is less than the total number of students in the (i + 1)th group, for all groups (except the last).

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 <= 105

• 1 <= grades[i] <= 105

## Solution (Java, C++, Python)

• class Solution {
public int maximumGroups(int[] grades) {
int len = grades.length;
return (int) (-1 + Math.sqrt(1D + 8 * len)) / 2;
}
}

############

class Solution {
public int maximumGroups(int[] grades) {
int n = grades.length;
int l = 0, r = n;
while (l < r) {
int mid = (l + r + 1) >> 1;
if (1L * mid * mid + mid > n * 2L) {
r = mid - 1;
} else {
l = mid;
}
}
return l;
}
}

• class Solution:
def maximumGroups(self, grades: List[int]) -> int:
ans = 1
prev = [1, grades[0]]
curr = [0, 0]
for v in grades[1:]:
curr[0] += 1
curr[1] += v
if prev[0] < curr[0] and prev[1] < curr[1]:
prev = curr
curr = [0, 0]
ans += 1
return ans

############

# 2358. Maximum Number of Groups Entering a Competition
# https://leetcode.com/problems/maximum-number-of-groups-entering-a-competition/

class Solution:
def maximumGroups(self, grades: List[int]) -> int:
res = 1
currSize = 0
size = 2

for i in range(1, n):
currSize += 1

if currSize == size:
res = currSize
size += 1
currSize = 0

return res


• class Solution {
public:
int maximumGroups(vector<int>& grades) {
int n = grades.size();
int l = 0, r = n;
while (l < r) {
int mid = (l + r + 1) >> 1;
if (1LL * mid * mid + mid > n * 2LL) {
r = mid - 1;
} else {
l = mid;
}
}
return l;
}
};

• func maximumGroups(grades []int) int {
return sort.Search(n, func(k int) bool {
k++
return k*k+k > n*2
})
}

• function maximumGroups(grades: number[]): number {
const n = grades.length;
let l = 1;
let r = n;
while (l < r) {
const mid = (l + r + 1) >> 1;
if (mid * mid + mid > n * 2) {
r = mid - 1;
} else {
l = mid;
}
}
return l;
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).