# 2790. Maximum Number of Groups With Increasing Length

## Description

You are given a 0-indexed array usageLimits of length n.

Your task is to create groups using numbers from 0 to n - 1, ensuring that each number, i, is used no more than usageLimits[i] times in total across all groups. You must also satisfy the following conditions:

• Each group must consist of distinct numbers, meaning that no duplicate numbers are allowed within a single group.
• Each group (except the first one) must have a length strictly greater than the previous group.

Return an integer denoting the maximum number of groups you can create while satisfying these conditions.

Example 1:

Input: usageLimits = [1,2,5]
Output: 3
Explanation: In this example, we can use 0 at most once, 1 at most twice, and 2 at most five times.
One way of creating the maximum number of groups while satisfying the conditions is:
Group 1 contains the number .
Group 2 contains the numbers [1,2].
Group 3 contains the numbers [0,1,2].
It can be shown that the maximum number of groups is 3.
So, the output is 3. 

Example 2:

Input: usageLimits = [2,1,2]
Output: 2
Explanation: In this example, we can use 0 at most twice, 1 at most once, and 2 at most twice.
One way of creating the maximum number of groups while satisfying the conditions is:
Group 1 contains the number .
Group 2 contains the numbers [1,2].
It can be shown that the maximum number of groups is 2.
So, the output is 2.


Example 3:

Input: usageLimits = [1,1]
Output: 1
Explanation: In this example, we can use both 0 and 1 at most once.
One way of creating the maximum number of groups while satisfying the conditions is:
Group 1 contains the number .
It can be shown that the maximum number of groups is 1.
So, the output is 1.


Constraints:

• 1 <= usageLimits.length <= 105
• 1 <= usageLimits[i] <= 109

## Solutions

• class Solution {
public int maxIncreasingGroups(List<Integer> usageLimits) {
Collections.sort(usageLimits);
int k = 0;
long s = 0;
for (int x : usageLimits) {
s += x;
if (s > k) {
++k;
s -= k;
}
}
return k;
}
}

• class Solution {
public:
int maxIncreasingGroups(vector<int>& usageLimits) {
sort(usageLimits.begin(), usageLimits.end());
int k = 0;
long long s = 0;
for (int x : usageLimits) {
s += x;
if (s > k) {
++k;
s -= k;
}
}
return k;
}
};

• class Solution:
def maxIncreasingGroups(self, usageLimits: List[int]) -> int:
usageLimits.sort()
k = s = 0
for x in usageLimits:
s += x
if s > k:
k += 1
s -= k
return k


• func maxIncreasingGroups(usageLimits []int) int {
sort.Ints(usageLimits)
s, k := 0, 0
for _, x := range usageLimits {
s += x
if s > k {
k++
s -= k
}
}
return k
}

• function maxIncreasingGroups(usageLimits: number[]): number {
usageLimits.sort((a, b) => a - b);
let k = 0;
let s = 0;
for (const x of usageLimits) {
s += x;
if (s > k) {
++k;
s -= k;
}
}
return k;
}