Leetcode 2790: Maximum Number of Groups With Increasing Length

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 <= 10^5
• 1 <= usageLimits[i] <= 10^9

Problem Analysis:

Assume we can create the maximum number of groups is n, then usageLimits should have the length ≥ n, and total times should ≥ 1 + 2 + … + n = n*(n+1)/2.

For avoid the case like [1, 1, 5], which satisfy above conditions, but the maximum number of groups is 2, not 3. So we can sort the usageLimits, and iterate check the conditions, so when i = 1, the subarray is [1, 1], which get the result n = 1, so when i = 2, the array [1, 1, 5] get the result n = 2.

Solution

class Solution {
public:
    int maxIncreasingGroups(vector<int>& usageLimits) {
        long long res{0};
        long long total{0};
        sort(usageLimits.begin(), usageLimits.end());
        for (const auto& item : usageLimits) {
            total += item;
            if (total * 2 >= (res + 1) * (res + 2))
                ++res;
        }
        
        return res;
    }
};

Time complexity is O(nlogn)

Space complexity is O(1)