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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 [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 <= 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, n = 0, len(usageLimits)
        for i in range(n):
            if usageLimits[i] > k:
                k += 1
                usageLimits[i] -= k
            if i + 1 < n:
                usageLimits[i + 1] += usageLimits[i]
        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;
}

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