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2790. Maximum Number of Groups With Increasing Length
Description
You are given a 0indexed 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 [2]. 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 [0]. 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 [0]. It can be shown that the maximum number of groups is 1. So, the output is 1.
Constraints:
1 <= usageLimits.length <= 10^{5}
1 <= usageLimits[i] <= 10^{9}
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; }