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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 = 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;
    }
    
    

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